NCERT Class 10 Maths Solutions: Introduction to Trigonometry

Question:

(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.

A. 0

B. 1

C. 2

D. -1

E. none of these

Concept in a Minute:

Trigonometric Identities: Specifically, the fundamental Pythagorean identities and relationships between trigonometric functions (tan, cot, sec, cosec) and sin/cos.
Algebraic Manipulation: Expanding and simplifying algebraic expressions, particularly those involving sums and products.

Explanation:

Let’s simplify the given expression: (1 + tan θ + sec θ) (1 + cot θ − cosec θ)

First, express tan θ, cot θ, sec θ, and cosec θ in terms of sin θ and cos θ:
tan θ = sin θ / cos θ
cot θ = cos θ / sin θ
sec θ = 1 / cos θ
cosec θ = 1 / sin θ

Substitute these into the expression:
(1 + sin θ / cos θ + 1 / cos θ) (1 + cos θ / sin θ − 1 / sin θ)

Combine terms within each parenthesis by finding a common denominator:
((cos θ + sin θ + 1) / cos θ) ((sin θ + cos θ – 1) / sin θ)

Now, multiply the two fractions:
((cos θ + sin θ + 1) (sin θ + cos θ – 1)) / (cos θ sin θ)

Rearrange the terms in the numerator to make it easier to see a pattern. Let (cos θ + sin θ) = A and 1 = B. The numerator looks like (A + B)(A – B), which is a difference of squares (A² – B²).
Numerator = ((cos θ + sin θ)² – 1²)
Numerator = (cos² θ + sin² θ + 2 sin θ cos θ) – 1

Using the identity cos² θ + sin² θ = 1:
Numerator = (1 + 2 sin θ cos θ) – 1
Numerator = 2 sin θ cos θ

Now, substitute this back into the multiplied expression:
(2 sin θ cos θ) / (cos θ sin θ)

Cancel out the common terms (sin θ cos θ) from the numerator and denominator:
2

Therefore, the expression simplifies to 2.

The final answer is $\boxed{2}$.

Question:

$2⁢tan⁡30°1+tan2⁡30°$ = ______.

A. sin 60°

B. cos 60°

C. tan 60°

D. sin 30°

Concept in a Minute:

Trigonometric identities, specifically the double angle formula for tangent, and the values of trigonometric ratios for standard angles (30° and 60°).

Explanation:

The given expression is $\frac{2\tan30°}{1+\tan^230°}$.
We know the value of $\tan30° = \frac{1}{\sqrt{3}}$.
Substitute this value into the expression:
$\frac{2(\frac{1}{\sqrt{3}})}{1+(\frac{1}{\sqrt{3}})^2} = \frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$
$= \frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$
$= \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}}$
To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$:
$\frac{3\sqrt{3}}{2\sqrt{3} \times \sqrt{3}} = \frac{3\sqrt{3}}{2 \times 3} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}$

Now, let’s check the options:
A. $\sin60° = \frac{\sqrt{3}}{2}$
B. $\cos60° = \frac{1}{2}$
C. $\tan60° = \sqrt{3}$
D. $\sin30° = \frac{1}{2}$

The calculated value matches $\sin60°$.

Alternatively, we can use the double angle identity for tangent:
$\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$
However, the given expression has a plus sign in the denominator, not a minus sign.
The expression $\frac{2\tan\theta}{1+\tan^2\theta}$ is equal to $\sin(2\theta)$.
Let $\theta = 30°$.
Then, $\frac{2\tan30°}{1+\tan^230°} = \sin(2 \times 30°) = \sin60°$.
We know that $\sin60° = \frac{\sqrt{3}}{2}$.

Final Answer: The final answer is $\boxed{\text{sin 60°}}$

Question:

State whether the following is true or false. Justify your answer.

The value of sinθ increases as θ increases.

A. True

B. False

Concept in a Minute:

Understanding the behavior of trigonometric functions, specifically the sine function, in the range of angles from 0 to 90 degrees. Knowledge of the unit circle or the graph of the sine function is essential.

Explanation:

The statement “The value of sinθ increases as θ increases” is true for angles in the first quadrant (0° to 90°).
To justify this, consider the unit circle. For an angle θ in the first quadrant, sinθ represents the y-coordinate of the point where the terminal side of the angle intersects the unit circle. As θ increases from 0° to 90°, the point on the unit circle moves upwards along the arc. Consequently, the y-coordinate (which is sinθ) also increases.
We can see this with specific values:
sin(0°) = 0
sin(30°) = 1/2 = 0.5
sin(45°) = 1/√2 ≈ 0.707
sin(60°) = √3/2 ≈ 0.866
sin(90°) = 1
As θ increases from 0° to 90°, the value of sinθ increases from 0 to 1.

However, it’s important to note that this statement is not true for all values of θ. For angles beyond 90°, the sine function starts to decrease. For example, sin(120°) = √3/2, which is less than sin(90°). But given the context of high school NCERT questions, this statement is typically considered in the context of acute angles where the function is monotonically increasing. Assuming the question implicitly refers to the first quadrant or angles where sinθ is increasing.

Therefore, the statement is True.

Answer: A

Question:

State whether the following is true or false. Justify your answer.

The value of cos θ increases as θ increases.

A. True

B. False

Concept in a Minute:

Understanding the behavior of the cosine function, specifically its values and how they change with respect to the angle θ. This involves recalling or visualizing the unit circle and the graph of the cosine function.

Explanation:

The statement “The value of cos θ increases as θ increases” is False.

To justify this, let’s consider the behavior of the cosine function in the first quadrant (0° to 90° or 0 to π/2 radians).
When θ = 0°, cos θ = 1.
When θ = 90°, cos θ = 0.
As θ increases from 0° to 90°, the value of cos θ decreases from 1 to 0.

Let’s consider the interval from 90° to 180° (or π/2 to π radians).
When θ = 90°, cos θ = 0.
When θ = 180°, cos θ = -1.
As θ increases from 90° to 180°, the value of cos θ decreases from 0 to -1.

In general, for angles between 0° and 180° (0 to π radians), the cosine function is decreasing. For angles between 180° and 360° (π to 2π radians), the cosine function increases. However, the statement claims it *always* increases as θ increases, which is not true for all values of θ. The cosine function is periodic and exhibits both increasing and decreasing intervals.

The graph of y = cos θ starts at its maximum value (1) at θ = 0, decreases to its minimum value (-1) at θ = π, and then increases back to its maximum value (1) at θ = 2π. Therefore, the value of cos θ does not always increase as θ increases.

The final answer is $\boxed{B}$.

Question:

$1+tan2⁡𝐴1+cot2⁡𝐴$ = ______.

A. sec 2 A

B. −1

C. cot 2 A

D. tan 2 A

Concept in a Minute:

Trigonometric identities, specifically the Pythagorean identities: $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \csc^2 A$. Also, the relationship between tangent and cotangent: $\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$.

Explanation:

We are asked to simplify the expression $\frac{1+\tan^2 A}{1+\cot^2 A}$.
We can use the Pythagorean identities to replace the numerator and the denominator.
The identity $1 + \tan^2 A$ is equal to $\sec^2 A$.
The identity $1 + \cot^2 A$ is equal to $\csc^2 A$.
So, the expression becomes $\frac{\sec^2 A}{\csc^2 A}$.

Now, we can express $\sec A$ and $\csc A$ in terms of $\sin A$ and $\cos A$.
We know that $\sec A = \frac{1}{\cos A}$ and $\csc A = \frac{1}{\sin A}$.
Therefore, $\sec^2 A = \frac{1}{\cos^2 A}$ and $\csc^2 A = \frac{1}{\sin^2 A}$.

Substituting these into our expression:
$\frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}}$

To divide by a fraction, we multiply by its reciprocal:
$\frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1} = \frac{\sin^2 A}{\cos^2 A}$

We know that $\frac{\sin A}{\cos A} = \tan A$.
Therefore, $\frac{\sin^2 A}{\cos^2 A} = \left(\frac{\sin A}{\cos A}\right)^2 = \tan^2 A$.

Thus, $1+\tan2⁡𝐴1+cot2⁡𝐴$ = $\tan^2 A$.

The correct option is D.

Question:

(secA + tanA) (1 − sinA) = ______.

A. sec A

B. sin A

C. cosec A

D. cos A

Concept in a Minute:

Trigonometric Identities: Specifically, the Pythagorean identity (sin^2 A + cos^2 A = 1) and the definitions of secant and tangent in terms of sine and cosine (sec A = 1/cos A, tan A = sin A/cos A).

Explanation:

We are asked to simplify the expression (secA + tanA) (1 − sinA).
First, let’s express secA and tanA in terms of sinA and cosA.
We know that secA = 1/cosA and tanA = sinA/cosA.
Substitute these into the expression:
(1/cosA + sinA/cosA) (1 − sinA)

Combine the terms inside the first parenthesis:
((1 + sinA)/cosA) (1 − sinA)

Now, multiply the two fractions:
(1 + sinA)(1 − sinA) / cosA

The numerator is a difference of squares, (a+b)(a-b) = a^2 – b^2. Here, a=1 and b=sinA.
So, the numerator becomes 1^2 – sin^2 A = 1 – sin^2 A.

Using the Pythagorean identity, we know that sin^2 A + cos^2 A = 1.
Rearranging this identity, we get cos^2 A = 1 – sin^2 A.
Substitute this back into the numerator:
cos^2 A / cosA

Now, simplify the expression by canceling out one cosA from the numerator and the denominator:
cos A

Therefore, (secA + tanA) (1 − sinA) = cos A.

The correct option is D.

Question:

State whether the following are true or false. Justify your answer.

The value of tan A is always less than 1.

A. True

B. False

Concept in a Minute:

The question tests the understanding of the range of values for the tangent function (tan A) for angles in a right-angled triangle. Specifically, it relates to acute angles.

Explanation:

The statement “The value of tan A is always less than 1” is False.
In a right-angled triangle, tan A is defined as the ratio of the length of the side opposite to angle A to the length of the adjacent side (tan A = opposite/adjacent).
Consider a right-angled triangle where angle A is an acute angle (0° < A < 90°).
If angle A is less than 45°, the side opposite to A will be shorter than the adjacent side, so tan A will be less than 1. For example, if A = 30°, tan 30° = 1/√3 ≈ 0.577, which is less than 1.
However, if angle A is greater than 45° (but less than 90°), the side opposite to A will be longer than the adjacent side, so tan A will be greater than 1. For example, if A = 60°, tan 60° = √3 ≈ 1.732, which is greater than 1.
When A = 45°, tan 45° = 1.
As angle A approaches 90°, tan A approaches infinity.
Therefore, the value of tan A is not always less than 1.

Answer: B. False

Question:

State whether the following are true or false. Justify your answer.

sec A = $125$ for some value of angle A.

A. True

B. False

Concept in a Minute:

The range of the secant function. The secant of an angle A, denoted as sec A, is defined as 1/cos A. The cosine function has a range of [-1, 1]. Therefore, the secant function can only take values that are less than or equal to -1 or greater than or equal to 1.

Explanation:

The secant function is defined as the reciprocal of the cosine function, i.e., sec A = 1/cos A.
The range of the cosine function is [-1, 1]. This means that for any angle A, -1 ≤ cos A ≤ 1.
When we take the reciprocal of values within the range [-1, 1], we need to consider two cases:
Case 1: cos A is positive (0 < cos A ≤ 1). In this case, sec A = 1/cos A will be ≥ 1.
Case 2: cos A is negative (-1 ≤ cos A < 0). In this case, sec A = 1/cos A will be ≤ -1.
Therefore, the range of the secant function is (-∞, -1] ∪ [1, ∞).
The given statement is sec A = 125. Since 125 is greater than or equal to 1, it falls within the possible range of the secant function. Thus, there exists a value of angle A for which sec A = 125.

Final Answer: The final answer is $\boxed{True}$

Question:

sin 2A = 2 sin A is true when A = ______.

A. 0°

B. 30°

C. 45°

D. 60°

Concept in a Minute:

Trigonometric identity for sin 2A and evaluating trigonometric functions for standard angles.

Explanation:

The question asks for the value of angle A for which the equation sin 2A = 2 sin A is true.
We need to test each of the given options.

Option A: A = 0°
LHS: sin (2 * 0°) = sin 0° = 0
RHS: 2 sin 0° = 2 * 0 = 0
Since LHS = RHS, sin 2A = 2 sin A is true for A = 0°.

Option B: A = 30°
LHS: sin (2 * 30°) = sin 60° = √3 / 2
RHS: 2 sin 30° = 2 * (1 / 2) = 1
Since LHS ≠ RHS, sin 2A = 2 sin A is not true for A = 30°.

Option C: A = 45°
LHS: sin (2 * 45°) = sin 90° = 1
RHS: 2 sin 45° = 2 * (1 / √2) = √2
Since LHS ≠ RHS, sin 2A = 2 sin A is not true for A = 45°.

Option D: A = 60°
LHS: sin (2 * 60°) = sin 120° = sin (180° – 60°) = sin 60° = √3 / 2
RHS: 2 sin 60° = 2 * (√3 / 2) = √3
Since LHS ≠ RHS, sin 2A = 2 sin A is not true for A = 60°.

Therefore, the equation sin 2A = 2 sin A is true when A = 0°.

The final answer is $\boxed{0°}$.

Question:

$1−tan2⁡45°1+tan2⁡45°$ = ______

A. tan 90°

B. 1

C. sin 45°

D. 0

Concept in a Minute:

Trigonometric identity for $\cos(2\theta)$: $\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}$.
Value of $\tan(45^\circ)$: $\tan(45^\circ) = 1$.

Explanation:

The given expression is $\frac{1-\tan^2(45^\circ)}{1+\tan^2(45^\circ)}$.
We know that $\tan(45^\circ) = 1$.
Substitute this value into the expression:
$\frac{1-(1)^2}{1+(1)^2} = \frac{1-1}{1+1} = \frac{0}{2} = 0$.

Alternatively, we can use the trigonometric identity $\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}$.
In this case, $\theta = 45^\circ$.
So, $\frac{1-\tan^2(45^\circ)}{1+\tan^2(45^\circ)} = \cos(2 \times 45^\circ) = \cos(90^\circ)$.
We know that $\cos(90^\circ) = 0$.

Therefore, the value of the expression is 0.

Comparing with the given options:
A. $\tan 90^\circ$ is undefined.
B. $1$.
C. $\sin 45^\circ = \frac{1}{\sqrt{2}}$.
D. $0$.

The correct option is D.

The final answer is $\boxed{0}$.

Question:

State whether the following are true or false. Justify your answer.

sin θ = $43$, for some angle θ.

A. True

B. False

Concept in a Minute:

The sine of an angle in a right-angled triangle is defined as the ratio of the length of the side opposite to the angle to the length of the hypotenuse. In any right-angled triangle, the hypotenuse is always the longest side. Therefore, the sine of an angle must always be less than or equal to 1. Mathematically, for any angle θ, -1 ≤ sin θ ≤ 1.

Explanation:

The question asks whether sin θ = 4/3 is true for some angle θ.
We know that the value of the sine function for any real angle θ lies in the range [-1, 1]. This means that the sine of an angle can never be greater than 1 or less than -1.
In this case, the given value is 4/3, which is approximately 1.33.
Since 4/3 > 1, it is impossible for sin θ to be equal to 4/3 for any real angle θ.
Therefore, the statement is false.

The final answer is $\boxed{False}$.

Question:

$2⁢tan⁡30°1−tan2⁡30°$ = ______.

A. cos 60°

B. sin 60°

C. tan 60°

D. sin 30°

Concept in a Minute:

Trigonometric identities, specifically the double angle formula for tangent: tan(2θ) = (2tanθ) / (1 – tan^2θ). Also, knowledge of the values of trigonometric functions for standard angles like 30° and 60°.

Explanation:

The given expression is $2⁢tan⁡30°1−tan2⁡30°$.
This expression is in the form of the double angle identity for tangent, which is tan(2θ) = (2tanθ) / (1 – tan^2θ).
In this case, θ = 30°.
So, $2⁢tan⁡30°1−tan2⁡30°$ = tan(2 * 30°) = tan(60°).

Now let’s evaluate the options:
A. cos 60° = 1/2
B. sin 60° = √3/2
C. tan 60° = √3
D. sin 30° = 1/2

Comparing our result tan(60°) with the options, we see that option C matches.

The final answer is $\boxed{tan 60°}$.

Question:

State whether the following is true or false. Justify your answer.

sinθ = cosθ for all values of θ.

A. True

B. False

Concept in a Minute:

Trigonometric Identities: Understanding the values of sine and cosine functions for different angles is crucial. Specifically, knowing when sinθ and cosθ are equal is key.

Explanation:

The statement “sinθ = cosθ for all values of θ” is false.
Trigonometric functions sine (sin) and cosine (cos) have different values for most angles. They are equal only for specific angles. For example, sin(45°) = cos(45°) = 1/√2. However, for other angles like θ = 0°, sin(0°) = 0 and cos(0°) = 1, which are not equal. Similarly, for θ = 30°, sin(30°) = 1/2 and cos(30°) = √3/2, which are not equal. Therefore, the statement is false.

The final answer is $\boxed{B}$.

Question:

State whether the following are true or false. Justify your answer.

cot A is not defined for A = 0°.

A. True

B. False

Concept in a Minute:

Trigonometric function cotangent (cot A) is defined as the ratio of cosine of an angle to the sine of that angle, i.e., cot A = cos A / sin A. It is undefined when the denominator (sin A) is zero.

Explanation:

We need to determine if cot A is undefined for A = 0°.
The definition of cotangent is: cot A = cos A / sin A.
Let’s evaluate sin A for A = 0°.
We know that sin(0°) = 0.
Since the denominator (sin A) is 0 when A = 0°, the expression cos(0°) / sin(0°) becomes cos(0°) / 0, which is undefined.
Therefore, cot A is not defined for A = 0°.

The statement “cot A is not defined for A = 0°” is true.

The final answer is $\boxed{A}$.

Question:

State whether the following are true or false. Justify your answer.

cos A is the abbreviation used for the cosecant of angle A.

A. True

B. False

Concept in a Minute:

The question tests knowledge of trigonometric function abbreviations. Specifically, it requires understanding that ‘cos A’ is the abbreviation for the cosine of angle A, not the cosecant.

Explanation:

The statement “cos A is the abbreviation used for the cosecant of angle A” is false.
In trigonometry:
‘cos A’ is the standard abbreviation for the cosine of angle A.
‘cosec A’ or ‘csc A’ is the standard abbreviation for the cosecant of angle A.
The question incorrectly equates the abbreviation for cosine with the cosecant.

Therefore, the correct answer is False.

Final Answer: The final answer is $\boxed{B}$

Question:

State whether the following are true or false. Justify your answer.

cot A is the product of cot and A.

A. True

B. False

Concept in a Minute:

Understanding trigonometric functions and their notation.

Explanation:

The expression ‘cot A’ represents the cotangent of an angle A. It is a single trigonometric function, not a product of two separate entities, ‘cot’ and ‘A’. In mathematics, when we write a trigonometric function followed by a variable (like sin x, cos θ, tan B, cot A, sec P, cosec Q), it signifies the function applied to that angle. There is no multiplication happening between the name of the function and the angle. Therefore, the statement that ‘cot A is the product of cot and A’ is false.

Answer:
B. False

Question:

State whether the following is true or false. Justify your answer.

sin (A + B) = sin A + sin B

A. True

B. False

Concept in a Minute:

Trigonometric identities, specifically the sine addition formula.

Explanation:

The statement sin (A + B) = sin A + sin B is false. The correct trigonometric identity for the sine of the sum of two angles is sin (A + B) = sin A cos B + cos A sin B. This can be demonstrated with a simple example. Let A = 30 degrees and B = 60 degrees.
sin (A + B) = sin (30 + 60) = sin (90) = 1.
sin A + sin B = sin 30 + sin 60 = 1/2 + sqrt(3)/2 = (1 + sqrt(3))/2.
Since 1 is not equal to (1 + sqrt(3))/2, the given statement is false.

Answer: B. False

Question:

9 sec² A − 9 tan² A = ______.

A. 1

B. 9

C. 8

D. 0

Concept in a Minute:

Trigonometric Identities. Specifically, the Pythagorean identity relating secant and tangent: sec^2(theta) – tan^2(theta) = 1.

Explanation:

The given expression is 9 sec^2 A – 9 tan^2 A.
We can factor out the common factor of 9 from both terms:
9 (sec^2 A – tan^2 A).
Now, we know the fundamental trigonometric identity: sec^2 A – tan^2 A = 1.
Substituting this identity into our expression, we get:
9 (1).
Therefore, 9 sec^2 A – 9 tan^2 A = 9.

The correct option is B.

Question:

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

(cosec A – sin A) (sec A – cos A) = $1tan⁡𝐴+cot⁡𝐴$

[Hint: Simplify LHS and RHS separately.]

Concept in a Minute:

Trigonometric identities, specifically the fundamental identities involving sine, cosine, tangent, cotangent, secant, and cosecant. The ability to express trigonometric functions in terms of sine and cosine is crucial. Algebraic manipulation of these expressions is also required.

Explanation:

We need to prove the identity (cosec A – sin A) (sec A – cos A) = 1/(tan A + cot A). We will simplify the Left Hand Side (LHS) and the Right Hand Side (RHS) separately and show that they are equal.

Step 1: Simplify the LHS.
LHS = (cosec A – sin A) (sec A – cos A)
Recall that cosec A = 1/sin A and sec A = 1/cos A. Substitute these into the expression.
LHS = (1/sin A – sin A) (1/cos A – cos A)
To combine the terms within each parenthesis, find a common denominator.
LHS = ((1 – sin^2 A)/sin A) ((1 – cos^2 A)/cos A)
Using the Pythagorean identity sin^2 A + cos^2 A = 1, we know that 1 – sin^2 A = cos^2 A and 1 – cos^2 A = sin^2 A.
LHS = (cos^2 A / sin A) (sin^2 A / cos A)
Now, multiply the fractions.
LHS = (cos^2 A * sin^2 A) / (sin A * cos A)
Cancel out one sin A and one cos A from the numerator and denominator.
LHS = cos A * sin A

Step 2: Simplify the RHS.
RHS = 1/(tan A + cot A)
Recall that tan A = sin A/cos A and cot A = cos A/sin A. Substitute these into the expression.
RHS = 1/(sin A/cos A + cos A/sin A)
To combine the terms in the denominator, find a common denominator.
RHS = 1/((sin^2 A + cos^2 A) / (cos A * sin A))
Using the Pythagorean identity sin^2 A + cos^2 A = 1.
RHS = 1/(1 / (cos A * sin A))
When you divide 1 by a fraction, you multiply by its reciprocal.
RHS = cos A * sin A

Step 3: Compare LHS and RHS.
We found that LHS = cos A * sin A and RHS = cos A * sin A.
Since LHS = RHS, the identity is proven.

The final answer is $\boxed{(cosec A – sin A) (sec A – cos A) = 1/(tan A+cot A)}$.

Question:

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

(sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

Concept in a Minute:

The problem requires proving a trigonometric identity. The key concepts involved are:
1. Squaring binomials: (a + b)^2 = a^2 + 2ab + b^2
2. Fundamental trigonometric identities:
* cosec A = 1/sin A
* sec A = 1/cos A
* tan A = sin A/cos A
* cot A = cos A/sin A
* sin^2 A + cos^2 A = 1
* 1 + tan^2 A = sec^2 A
* 1 + cot^2 A = cosec^2 A

Explanation:

To prove the given identity, we will start by expanding the terms on the left-hand side (LHS) using the binomial expansion (a + b)^2 = a^2 + 2ab + b^2.

LHS = (sin A + cosec A)^2 + (cos A + sec A)^2

Expand the first term:
(sin A + cosec A)^2 = sin^2 A + 2(sin A)(cosec A) + cosec^2 A
Since cosec A = 1/sin A, the middle term becomes 2(sin A)(1/sin A) = 2.
So, (sin A + cosec A)^2 = sin^2 A + 2 + cosec^2 A

Expand the second term:
(cos A + sec A)^2 = cos^2 A + 2(cos A)(sec A) + sec^2 A
Since sec A = 1/cos A, the middle term becomes 2(cos A)(1/cos A) = 2.
So, (cos A + sec A)^2 = cos^2 A + 2 + sec^2 A

Now, add the expanded terms of the LHS:
LHS = (sin^2 A + 2 + cosec^2 A) + (cos^2 A + 2 + sec^2 A)
LHS = sin^2 A + cos^2 A + 2 + 2 + cosec^2 A + sec^2 A

Using the identity sin^2 A + cos^2 A = 1:
LHS = 1 + 4 + cosec^2 A + sec^2 A
LHS = 5 + cosec^2 A + sec^2 A

Now, we need to express cosec^2 A and sec^2 A in terms of tan^2 A and cot^2 A.
We know the identities:
cosec^2 A = 1 + cot^2 A
sec^2 A = 1 + tan^2 A

Substitute these into the LHS:
LHS = 5 + (1 + cot^2 A) + (1 + tan^2 A)
LHS = 5 + 1 + cot^2 A + 1 + tan^2 A
LHS = 7 + tan^2 A + cot^2 A

This is the RHS of the given identity.
Therefore, LHS = RHS, and the identity is proven.

Question:

In ΔABC right angled at B, AB = 24 cm, BC = 7 m. Determine:

sin C, cos C

Concept in a Minute:

Trigonometric Ratios in a Right-Angled Triangle: Sine (sin) and Cosine (cos) are defined in terms of the sides of a right-angled triangle. For an angle C in a right-angled triangle ABC (right-angled at B):
sin C = (Length of the side opposite to angle C) / (Length of the hypotenuse)
cos C = (Length of the side adjacent to angle C) / (Length of the hypotenuse)
Pythagorean Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. AC² = AB² + BC²

Explanation:

The problem asks us to find the values of sin C and cos C in a right-angled triangle ABC, where the right angle is at B. We are given the lengths of two sides: AB = 24 cm and BC = 7 cm.

First, we need to identify the sides of the triangle relative to angle C.
The side opposite to angle C is AB.
The side adjacent to angle C is BC.
The hypotenuse is the side opposite to the right angle B, which is AC.

We are given the lengths of the two legs (AB and BC), but we need the length of the hypotenuse (AC) to calculate sin C and cos C. We can find the length of the hypotenuse using the Pythagorean theorem.

Pythagorean Theorem:
AC² = AB² + BC²
AC² = (24 cm)² + (7 cm)²
AC² = 576 cm² + 49 cm²
AC² = 625 cm²
AC = √625 cm²
AC = 25 cm

Now that we have the lengths of all three sides, we can calculate sin C and cos C.

To find sin C:
sin C = (Opposite side to C) / (Hypotenuse)
sin C = AB / AC
sin C = 24 cm / 25 cm
sin C = 24/25

To find cos C:
cos C = (Adjacent side to C) / (Hypotenuse)
cos C = BC / AC
cos C = 7 cm / 25 cm
cos C = 7/25

Therefore, sin C = 24/25 and cos C = 7/25.

Question:

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Concept in a Minute:

The question utilizes the property of the cosine function in trigonometry. Specifically, it relies on the fact that for acute angles (angles between 0 and 90 degrees, exclusive), the cosine function is a one-to-one function. This means that if the cosine of two acute angles is equal, then those angles must also be equal.

Explanation:

We are given that ∠A and ∠B are acute angles and cos A = cos B.
An acute angle is an angle that measures less than 90 degrees.
In the range of acute angles (0° < θ < 90°), the cosine function, cos(θ), is strictly decreasing. This means that as the angle increases, its cosine value decreases.
Because the cosine function is strictly decreasing for acute angles, it is a one-to-one function in this domain. A function is one-to-one if each output (in this case, the cosine value) corresponds to exactly one input (the angle).
Therefore, if cos A = cos B for acute angles A and B, it directly implies that the angles themselves must be equal, i.e., ∠A = ∠B.
This can be visualized by looking at the graph of the cosine function between 0° and 90°. Within this interval, no two different angles will have the same cosine value.

To formally show this:
Consider the inverse cosine function, arccos (or cos⁻¹).
If cos A = cos B, we can apply the inverse cosine function to both sides:
cos⁻¹(cos A) = cos⁻¹(cos B)
Since A is an acute angle, cos⁻¹(cos A) = A.
Similarly, since B is an acute angle, cos⁻¹(cos B) = B.
Therefore, A = B.
Thus, we have shown that if ∠A and ∠B are acute angles such that cos A = cos B, then ∠A = ∠B.

Question:

If tan (A + B) = $\sqrt{3}$ and tan (A – B) = $1\sqrt{3}$; 0° < A + B ≤ 90°; A > B, find A and B.

Concept in a Minute:

Trigonometric Ratios of Standard Angles: Knowledge of tangent values for standard angles (0°, 30°, 45°, 60°, 90°) is crucial.
Properties of Inverse Trigonometric Functions: Understanding that if tan(x) = tan(y), then x = y + n*180°, where n is an integer.
Solving Simultaneous Equations: Ability to solve a system of two linear equations with two variables.

Explanation:

We are given two equations involving the tangent of sums and differences of angles A and B:
1. tan (A + B) = $\sqrt{3}$
2. tan (A – B) = $1\sqrt{3}$

From the knowledge of standard trigonometric angles, we know that tan(60°) = $\sqrt{3}$ and tan(30°) = $\frac{1}{\sqrt{3}}$.

Using the first equation, tan (A + B) = $\sqrt{3}$:
Since tan(60°) = $\sqrt{3}$, we can write:
A + B = 60° (Equation 1)
The condition 0° < A + B ≤ 90° is satisfied as 60° falls within this range.

Using the second equation, tan (A – B) = $\frac{1}{\sqrt{3}}$:
Since tan(30°) = $\frac{1}{\sqrt{3}}$, we can write:
A – B = 30° (Equation 2)
The condition A > B is implicitly satisfied here, as A – B is a positive value (30°).

Now we have a system of two linear equations with two variables, A and B:
Equation 1: A + B = 60°
Equation 2: A – B = 30°

To solve for A and B, we can add Equation 1 and Equation 2:
(A + B) + (A – B) = 60° + 30°
2A = 90°
A = $\frac{90°}{2}$
A = 45°

Now substitute the value of A into Equation 1 to find B:
45° + B = 60°
B = 60° – 45°
B = 15°

We can check if the condition A > B is met: 45° > 15°, which is true.

Therefore, the values of A and B are 45° and 15° respectively.

Question:

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

$(cos⁡𝑒⁢𝑐θ–cot⁡θ)2=1−cos⁡𝜃1+cos⁡𝜃$

Concept in a Minute:

Trigonometric identities, specifically those involving cosecant and cotangent, and the manipulation of algebraic expressions with square roots. Key identities that might be useful are: cosec θ = 1/sin θ, cot θ = cos θ/sin θ, and the Pythagorean identity sin² θ + cos² θ = 1.

Explanation:

We need to prove the identity $(cos⁡𝑒⁢𝑐θ–cot⁡θ)2=1−cos⁡𝜃1+cos⁡𝜃$.
Let’s start with the Left Hand Side (LHS) and try to transform it into the Right Hand Side (RHS).

LHS = $(cos⁡𝑒⁢𝑐θ–cot⁡θ)2$

We know that cosec θ = 1/sin θ and cot θ = cos θ/sin θ.
Substitute these into the LHS:
LHS = $(\frac{1}{sin⁡θ} – \frac{cos⁡θ}{sin⁡θ})2$

Combine the terms inside the bracket since they have a common denominator:
LHS = $(\frac{1-cos⁡θ}{sin⁡θ})2$

Now, square the numerator and the denominator separately:
LHS = $\frac{(1-cos⁡θ)2}{sin^2⁡θ}$

We know the Pythagorean identity: sin² θ + cos² θ = 1.
From this, we can express sin² θ as 1 – cos² θ.
Substitute this into the denominator:
LHS = $\frac{(1-cos⁡θ)2}{1-cos^2⁡θ}$

The denominator is in the form of a² – b², where a=1 and b=cos θ. We can factor it as (a-b)(a+b).
So, 1 – cos² θ = (1 – cos θ)(1 + cos θ).
Substitute this factorization into the denominator:
LHS = $\frac{(1-cos⁡θ)2}{(1-cos⁡θ)(1+cos⁡θ)}$

Now, we can cancel out one of the (1 – cos θ) terms from the numerator and the denominator:
LHS = $\frac{1-cos⁡θ}{1+cos⁡θ}$

This is the Right Hand Side (RHS) of the identity.
Therefore, LHS = RHS.
The identity is proven.

Question:

In ΔABC, right angled at B. If tan A = $1\sqrt{3}$ , find the value of

sin A cos C + cos A sin C
cos A cos C − sin A sin C

Concept in a Minute:

Trigonometric Ratios in a Right-Angled Triangle: Understanding the definitions of tan, sin, and cos in terms of opposite, adjacent, and hypotenuse sides. Special angles: Recognizing the trigonometric values for standard angles like 30°, 45°, and 60°. Trigonometric Identities: Using the sum and difference identities for sine and cosine.

Explanation:

Given a right-angled triangle ABC, with the right angle at B. We are provided with tan A = $1\sqrt{3}$.

First, we need to determine the angle A. We know that tan 30° = $1\sqrt{3}$. Therefore, angle A = 30°.
Since the sum of angles in a triangle is 180° and angle B = 90°, and angle A = 30°, then angle C = 180° – 90° – 30° = 60°.

Now we can find the trigonometric values for angles A and C:
sin A = sin 30° = 1/2
cos A = cos 30° = $\sqrt{3}/2$
sin C = sin 60° = $\sqrt{3}/2$
cos C = cos 60° = 1/2

Now, let’s solve the given parts:

Part 1: sin A cos C + cos A sin C
Substitute the values:
(1/2) * (1/2) + ($\sqrt{3}/2$) * ($\sqrt{3}/2$)
= 1/4 + 3/4
= 4/4
= 1

Alternatively, we can recognize that sin A cos C + cos A sin C is the expansion of sin(A + C).
sin(A + C) = sin(30° + 60°) = sin(90°) = 1.

Part 2: cos A cos C − sin A sin C
Substitute the values:
($\sqrt{3}/2$) * (1/2) – (1/2) * ($\sqrt{3}/2$)
= $\sqrt{3}/4$ – $\sqrt{3}/4$
= 0

Alternatively, we can recognize that cos A cos C − sin A sin C is the expansion of cos(A + C).
cos(A + C) = cos(30° + 60°) = cos(90°) = 0.

Question:

Given 15 cot A = 8. Find sin A and sec A.

Concept in a Minute:

Trigonometric Ratios: Understanding the definitions of cotangent (cot), sine (sin), and secant (sec) in a right-angled triangle.
Pythagorean Theorem: The relationship between the sides of a right-angled triangle ($a^2 + b^2 = c^2$).

Explanation:

The question provides the value of cot A and asks us to find the values of sin A and sec A.

Step 1: Understand the given information.
We are given that 15 cot A = 8.

Step 2: Find the value of cot A.
Divide both sides of the equation by 15:
cot A = 8/15

Step 3: Relate cot A to the sides of a right-angled triangle.
In a right-angled triangle, cot A = Adjacent side / Opposite side.
So, we can consider the adjacent side to angle A as 8 units and the opposite side to angle A as 15 units.

Step 4: Use the Pythagorean theorem to find the hypotenuse.
Let the adjacent side be ‘b’ and the opposite side be ‘a’. Let the hypotenuse be ‘c’.
We have b = 8 and a = 15.
According to the Pythagorean theorem: $a^2 + b^2 = c^2$
$15^2 + 8^2 = c^2$
$225 + 64 = c^2$
$289 = c^2$
Taking the square root of both sides:
c = $\sqrt{289}$
c = 17

Step 5: Find sin A.
In a right-angled triangle, sin A = Opposite side / Hypotenuse.
sin A = a / c
sin A = 15 / 17

Step 6: Find sec A.
In a right-angled triangle, sec A = Hypotenuse / Adjacent side.
sec A = c / b
sec A = 17 / 8

Final Answer:
sin A = 15/17
sec A = 17/8

Question:

In ΔPQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Concept in a Minute:

Right-angled triangle trigonometry, Pythagorean theorem. Understanding trigonometric ratios (sin, cos, tan) in terms of sides of a right-angled triangle.

Explanation:

Let the sides of the right-angled triangle PQR be PQ, QR, and PR. We are given that the triangle is right-angled at Q.
According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In ΔPQR, PR is the hypotenuse.
So, PQ^2 + QR^2 = PR^2.

We are given:
1. PR + QR = 25 cm
2. PQ = 5 cm

From equation (1), we can express PR in terms of QR:
PR = 25 – QR

Now, substitute the values of PQ and PR into the Pythagorean theorem:
5^2 + QR^2 = (25 – QR)^2

25 + QR^2 = 625 – 50*QR + QR^2

Subtract QR^2 from both sides:
25 = 625 – 50*QR

Rearrange the equation to solve for QR:
50*QR = 625 – 25
50*QR = 600
QR = 600 / 50
QR = 12 cm

Now that we have QR, we can find PR using equation (1):
PR = 25 – QR
PR = 25 – 12
PR = 13 cm

So, the sides of the triangle are:
PQ = 5 cm (Perpendicular to angle P)
QR = 12 cm (Base to angle P)
PR = 13 cm (Hypotenuse)

Now, we can determine the trigonometric values for angle P:
sin P = Opposite side / Hypotenuse = PQ / PR = 5 / 13
cos P = Adjacent side / Hypotenuse = QR / PR = 12 / 13
tan P = Opposite side / Adjacent side = PQ / QR = 5 / 12

Question:

Evaluate the following:

$5⁢cos2⁡60°+4⁢sec2⁡30°−tan2⁡45°sin2⁡30°+cos2⁡30°$

Concept in a Minute:

Trigonometric Identities: This question requires knowledge of the values of trigonometric functions for standard angles (0°, 30°, 45°, 60°, 90°). Specifically, you’ll need to know cos 60°, sec 30°, tan 45°, sin 30°, and cos 30°. You will also need to apply basic arithmetic operations, including exponents and division.

Explanation:

The expression to evaluate is $5⁢cos2⁡60°+4⁢sec2⁡30°−tan2⁡45°sin2⁡30°+cos2⁡30°$.

First, let’s find the values of the trigonometric functions for the given angles:
cos 60° = 1/2
sec 30° = 1/cos 30° = 1/(√3/2) = 2/√3
tan 45° = 1
sin 30° = 1/2
cos 30° = √3/2

Now, substitute these values into the expression and square them where indicated:
cos²60° = (1/2)² = 1/4
sec²30° = (2/√3)² = 4/3
tan²45° = (1)² = 1
sin²30° = (1/2)² = 1/4
cos²30° = (√3/2)² = 3/4

Now, substitute these squared values back into the original expression:
5 * (1/4) + 4 * (4/3) – 1 * (1/4) + (3/4)

Perform the multiplications:
5/4 + 16/3 – 1/4 + 3/4

Combine the terms with the same denominator:
(5/4 – 1/4 + 3/4) + 16/3
(5 – 1 + 3)/4 + 16/3
7/4 + 16/3

To add these fractions, find a common denominator, which is 12:
(7 * 3)/(4 * 3) + (16 * 4)/(3 * 4)
21/12 + 64/12

Add the fractions:
(21 + 64)/12
85/12

The final answer is 85/12.

Question:

In ΔABC right angled at B, AB = 24 cm, BC = 7 m. Determine:

sin A, cos A

Concept in a Minute:

Trigonometric Ratios: In a right-angled triangle, the sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse. The cosine of an angle is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Explanation:

The question states that ΔABC is right-angled at B.
Given:
AB = 24 cm
BC = 7 cm

First, we need to find the length of the hypotenuse AC using the Pythagorean theorem.
AC^2 = AB^2 + BC^2
AC^2 = (24 cm)^2 + (7 cm)^2
AC^2 = 576 cm^2 + 49 cm^2
AC^2 = 625 cm^2
AC = sqrt(625 cm^2)
AC = 25 cm

Now we can determine sin A and cos A.

For sin A:
sin A = (Opposite side to angle A) / (Hypotenuse)
The side opposite to angle A is BC.
The hypotenuse is AC.
sin A = BC / AC
sin A = 7 cm / 25 cm
sin A = 7/25

For cos A:
cos A = (Adjacent side to angle A) / (Hypotenuse)
The side adjacent to angle A is AB.
The hypotenuse is AC.
cos A = AB / AC
cos A = 24 cm / 25 cm
cos A = 24/25

Therefore, sin A = 7/25 and cos A = 24/25.

Question:

Evaluate the following in the simplest form:

sin 60° cos 30° + cos 60° sin 30°

Concept in a Minute:

Trigonometric Identities: Specifically, the sine addition formula: sin(A + B) = sin A cos B + cos A sin B. Also, knowledge of the values of trigonometric functions for standard angles like 30°, 60° is essential.

Explanation:

The given expression is sin 60° cos 30° + cos 60° sin 30°.
We can recognize this as the expansion of the sine addition formula: sin(A + B) = sin A cos B + cos A sin B.
In this case, let A = 60° and B = 30°.
So, sin 60° cos 30° + cos 60° sin 30° = sin (60° + 30°).
Adding the angles inside the sine function: 60° + 30° = 90°.
Therefore, the expression simplifies to sin 90°.
The value of sin 90° is 1.

Alternatively, we can substitute the known values of the trigonometric functions:
sin 60° = √3/2
cos 30° = √3/2
cos 60° = 1/2
sin 30° = 1/2

Substituting these values into the expression:
(√3/2) * (√3/2) + (1/2) * (1/2)
= (√3 * √3) / (2 * 2) + (1 * 1) / (2 * 2)
= 3/4 + 1/4
= (3 + 1) / 4
= 4/4
= 1.

Both methods yield the same result.

Final Answer: The final answer is $\boxed{1}$

Question:

Given sec θ = $1312$, calculate all other trigonometric ratios.

Concept in a Minute:

The question requires understanding the definitions of trigonometric ratios in a right-angled triangle and the Pythagorean identity. We need to relate the given secant value to other ratios.

Explanation:

Given sec θ = 13/12.
We know that sec θ = hypotenuse / adjacent side.
So, let the hypotenuse be 13k and the adjacent side be 12k for some positive constant k.
Using the Pythagorean theorem, (hypotenuse)² = (adjacent side)² + (opposite side)².
(13k)² = (12k)² + (opposite side)²
169k² = 144k² + (opposite side)²
(opposite side)² = 169k² – 144k² = 25k²
opposite side = √(25k²) = 5k

Now we can find all other trigonometric ratios:

1. cos θ = adjacent side / hypotenuse = 12k / 13k = 12/13

2. sin θ = opposite side / hypotenuse = 5k / 13k = 5/13

3. tan θ = opposite side / adjacent side = 5k / 12k = 5/12

4. cosec θ = hypotenuse / opposite side = 13k / 5k = 13/5

5. cot θ = adjacent side / opposite side = 12k / 5k = 12/5

Question:

Evaluate the following:

$cos⁡45°sec⁡30°+cos⁡𝑒⁢𝑐30°$

Concept in a Minute:

Trigonometric Identities and Values: This question requires knowledge of the values of trigonometric functions for standard angles like 45° and 30°. Specifically, it involves cos 45°, sec 30°, and cosec 30°.

Explanation:

To evaluate the given expression $cos⁡45°sec⁡30°+cos⁡𝑒⁢𝑐30°$, we need to substitute the known trigonometric values for the angles 45° and 30°.

Step 1: Recall the standard trigonometric values:
cos 45° = 1/√2
sec 30° = 1/cos 30° = 1/(√3/2) = 2/√3
cosec 30° = 1/sin 30° = 1/(1/2) = 2

Step 2: Substitute these values into the expression:
(1/√2) * (2/√3) + 2

Step 3: Perform the multiplication:
2 / (√2 * √3) + 2
2 / √6 + 2

Step 4: Rationalize the denominator of the first term (optional but good practice):
(2/√6) * (√6/√6) + 2
2√6 / 6 + 2
√6 / 3 + 2

Step 5: Combine the terms to get the final answer:
(√6 + 6) / 3

Therefore, $cos⁡45°sec⁡30°+cos⁡𝑒⁢𝑐30° = (√6 + 6) / 3$.

Question:

Evaluate the following:

2tan²45° + cos²30° − sin²60°

Concept in a Minute:

This question requires the knowledge of trigonometric values of standard angles, specifically 45°, 30°, and 60°. It also involves basic arithmetic operations and understanding of exponents.

Explanation:

We need to evaluate the expression 2tan² 45° + cos² 30° − sin² 60°.
First, recall the standard trigonometric values:
tan 45° = 1
cos 30° = √3/2
sin 60° = √3/2

Now substitute these values into the expression:
2(tan 45°)² + (cos 30°)² − (sin 60°)²
= 2(1)² + (√3/2)² − (√3/2)²

Next, calculate the squares:
= 2(1) + (3/4) − (3/4)

Perform the multiplications and additions/subtractions:
= 2 + 3/4 − 3/4
= 2

Therefore, the value of the expression is 2.

Final Answer:
2

Question:

If cot θ = $78$, evaluate cot² θ.

Concept in a Minute:

The question involves trigonometric identities and basic algebraic manipulation. The key concept is understanding the meaning of cot θ and how to square a trigonometric function. If cot θ is given as a value, then cot² θ is simply the square of that value.

Explanation:

We are given that cot θ = 7/8.
We need to evaluate cot² θ.
The notation cot² θ means (cot θ)².
Therefore, to find cot² θ, we need to square the given value of cot θ.
cot² θ = (cot θ)²
cot² θ = (7/8)²
cot² θ = (7²)/(8²)
cot² θ = 49/64

Next Chapter: Pair of Linear Equations in Two Variables

Refer Introduction to Trigonometry Notes

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