Trigonometry: SSC CGL Exam

Ans: D

Explanation: This problem uses trigonometry. The ladder, the building, and the ground form a right triangle. The ladder is the hypotenuse (5 meters), the height the ladder reaches on the building is the opposite side to the 30-degree angle, and the distance on the ground is the adjacent side. We use the sine function: sin(angle) = opposite/hypotenuse. So, sin(30 degrees) = height / 5. sin(30) = 0.5. Therefore, height = 5 * 0.5 = 2.5 meters.

Ans: B

Explanation: This question utilizes the fundamental trigonometric identity: cos²θ + sin²θ = 1. In this case, θ = 60°. Therefore, cos²60° + sin²60° = 1.

Ans: C

Explanation: We can use trigonometry to solve this problem. The tree and its shadow form a right triangle. The height of the tree is the opposite side, and the length of the shadow is the adjacent side to the angle we want to find. We can use the tangent function: tan(angle) = opposite/adjacent. In this case, tan(angle) = 30 / (10√3) = 3 / √3 = √3. The angle whose tangent is √3 is 60 degrees.
Correct Option: C

Ans: A

Explanation: We can use trigonometric identities to simplify the expression.
Recall that:
1. csc(x) = 1/sin(x)
2. sec(x) = 1/cos(x)
3. csc(90 – x) = sec(x)
4. sec(90 – x) = csc(x)

The given expression is: csc(60° + A) – sec(30° – A) + csc(49°) / sec(41°)

Let’s simplify csc(60° + A) – sec(30° – A). We can rewrite sec(30° – A) as csc(90° – (30° – A)) = csc(60° + A). Therefore, csc(60° + A) – sec(30° – A) = csc(60° + A) – csc(60° + A) = 0.

Now, let’s simplify csc(49°) / sec(41°). We can rewrite sec(41°) as csc(90° – 41°) = csc(49°). Therefore, csc(49°) / sec(41°) = csc(49°) / csc(49°) = 1.

So, the original expression simplifies to: 0 + 1 = 1.

Correct Option: A

Ans: B

Explanation:
We can use the identity: tan(x) * tan(90-x) = 1.
We can rearrange the terms:
tan 10 * tan 80 = tan 10 * tan (90-10) = tan 10 * cot 10 = 1
tan 20 * tan 70 = tan 20 * tan (90-20) = tan 20 * cot 20 = 1
tan 30 * tan 60 = (1/sqrt(3)) * sqrt(3) = 1
tan 40 * tan 50 = tan 40 * tan (90-40) = tan 40 * cot 40 = 1

Therefore the expression is:
(tan 10 * tan 80) * (tan 20 * tan 70) * (tan 30 * tan 60) * (tan 40 * tan 50)
= 1 * 1 * 1 * 1 = 1

Ans: A

Explanation: We are given cos(θ) = 35/37. We need to find csc(θ). First, find sin(θ) using the Pythagorean identity: sin²(θ) + cos²(θ) = 1. Therefore, sin²(θ) = 1 – cos²(θ) = 1 – (35/37)² = 1 – 1225/1369 = (1369 – 1225)/1369 = 144/1369. Taking the square root gives sin(θ) = √(144/1369) = 12/37. Then, csc(θ) = 1/sin(θ) = 1 / (12/37) = 37/12.

Correct Option: A

Ans: C

Explanation:
Let sin(θ) = s and cos(θ) = c.
We are given that s⁶ + c⁶ = 1/3.
We know that s² + c² = 1.
We want to find the value of s*c.

We can rewrite s⁶ + c⁶ as (s²)³ + (c²)³.
Using the identity a³ + b³ = (a + b)(a² – ab + b²), we have:
s⁶ + c⁶ = (s² + c²)(s⁴ – s²c² + c⁴) = 1 * (s⁴ – s²c² + c⁴)
Since s⁶ + c⁶ = 1/3, we get:
s⁴ – s²c² + c⁴ = 1/3
Also, (s² + c²)² = s⁴ + 2s²c² + c⁴ = 1² = 1.
We can rewrite the first equation as:
(s⁴ + 2s²c² + c⁴) – 3s²c² = 1/3
1 – 3s²c² = 1/3
3s²c² = 1 – 1/3 = 2/3
s²c² = 2/9
sc = √(2/9) = √2 / 3

Correct Option: C

Ans: D

Explanation: We are given the equation $3 \sin^2 \theta – \cos \theta – 1 = 0$. We can rewrite this equation using the identity $\sin^2 \theta = 1 – \cos^2 \theta$. Substituting, we have:
$3(1 – \cos^2 \theta) – \cos \theta – 1 = 0$
$3 – 3\cos^2 \theta – \cos \theta – 1 = 0$
$3\cos^2 \theta + \cos \theta – 2 = 0$
Let $x = \cos \theta$. Then we have the quadratic equation $3x^2 + x – 2 = 0$.
We can factor this quadratic as $(3x – 2)(x + 1) = 0$.
Thus, the possible values for x (which is $\cos \theta$) are $x = \frac{2}{3}$ or $x = -1$.
Since $0 < \theta < 90^\circ$, we know that $0 < \cos \theta < 1$. Therefore, we must have $\cos \theta = \frac{2}{3}$.
Now, we can find $\sin \theta$ using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$\sin^2 \theta = 1 – \cos^2 \theta = 1 – \left(\frac{2}{3}\right)^2 = 1 – \frac{4}{9} = \frac{5}{9}$
Since $0 < \theta < 90^\circ$, we have $\sin \theta = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
Now, we can find $\cot \theta$ and $\csc \theta$:
$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}}$
$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{\sqrt{5}}{3}} = \frac{3}{\sqrt{5}}$
Finally, we can find $\cot \theta + \csc \theta$:
$\cot \theta + \csc \theta = \frac{2}{\sqrt{5}} + \frac{3}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$

Correct Option: D

Ans: D

Explanation: Since triangle DEF is a right-angled triangle with the right angle at E, we know that angle D + angle F = 90 degrees. Given that angle F is 45 degrees, angle D must also be 45 degrees. Thus, triangle DEF is a 45-45-90 triangle. We have: csc(F) = csc(45°) = 1/sin(45°) = 1/(1/√2) = √2. cot(D) = cot(45°) = 1. Therefore, csc(F) * cot(D) = √2 * 1 = √2.

Correct Option: D

Ans: C

Explanation:
We can use trigonometric identities to solve this problem.

First, recall the identity: cos(90 – x) = sin(x).
Therefore, cos(75°) = cos(90° – 15°) = sin(15°).

Now we can rewrite the expression:
cos²(15°) + cos²(75°) + cot²(45°) = cos²(15°) + sin²(15°) + cot²(45°)

Using the identity cos²(x) + sin²(x) = 1, we know cos²(15°) + sin²(15°) = 1.

Also, we know that cot(45°) = 1. So, cot²(45°) = 1² = 1.

Therefore, the original expression simplifies to:
1 + 1 = 2

Ans: B

Explanation: Let’s simplify the first part of the expression: ((sin(x+y) – 2sin(x) + sin(x-y))/(cos(x-y) + cos(x+y) – 2cos(x))). We can use the sum-to-product formulas:
sin(x+y) + sin(x-y) = 2sin(x)cos(y)
cos(x-y) + cos(x+y) = 2cos(x)cos(y)

So, the numerator becomes: 2sin(x)cos(y) – 2sin(x) = 2sin(x)(cos(y) – 1)
And the denominator becomes: 2cos(x)cos(y) – 2cos(x) = 2cos(x)(cos(y) – 1)

Therefore, the first part simplifies to: (2sin(x)(cos(y)-1)) / (2cos(x)(cos(y)-1)) = tan(x), if cos(y) != 1.

Now, let’s simplify the second part: ((sin(10x) – sin(8x))/(cos(10x) + cos(8x))). Using sum-to-product formulas:
sin(10x) – sin(8x) = 2cos((10x+8x)/2)sin((10x-8x)/2) = 2cos(9x)sin(x)
cos(10x) + cos(8x) = 2cos((10x+8x)/2)cos((10x-8x)/2) = 2cos(9x)cos(x)

Therefore, the second part simplifies to: (2cos(9x)sin(x)) / (2cos(9x)cos(x)) = tan(x), if cos(9x) != 0.

So the entire expression simplifies to: tan(x) * tan(x) = tan^2(x).

Correct Option: B

Ans: B

Explanation: We can use trigonometric identities to simplify this expression.

First, use the identity cos(A) – cos(B) = -2sin((A+B)/2)sin((A-B)/2) on the numerator:
cos 40° – cos 140° = -2sin((40+140)/2)sin((40-140)/2) = -2sin(90°)sin(-50°) = -2(1)(-sin 50°) = 2sin 50°

Next, use the identity sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2) on the denominator:
sin 80° + sin 20° = 2sin((80+20)/2)cos((80-20)/2) = 2sin(50°)cos(30°)

Now the original expression becomes:
(2sin 50°) / (2sin 50°cos 30°)

Simplifying, we get:
1 / cos 30°

Since cos 30° = √3/2, the expression equals:
1 / (√3/2) = 2/√3

Correct Option: B

Ans: B

Explanation: Let’s simplify the given expression step-by-step using trigonometric identities.
(tan θ + cot θ)(sec θ + tan θ)(1 – sin θ)

1. Express tan θ and cot 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

2. Substitute the expressions in the given expression:
[(sin θ / cos θ) + (cos θ / sin θ)] * [(1 / cos θ) + (sin θ / cos θ)] * (1 – sin θ)

3. Simplify the terms inside the brackets:
[(sin² θ + cos² θ) / (cos θ * sin θ)] * [(1 + sin θ) / cos θ] * (1 – sin θ)
Since sin² θ + cos² θ = 1,
[1 / (cos θ * sin θ)] * [(1 + sin θ) / cos θ] * (1 – sin θ)

4. Combine terms:
[(1 + sin θ)(1 – sin θ)] / [cos² θ * sin θ]
Using (a+b)(a-b) = a² – b²
(1 – sin² θ) / (cos² θ * sin θ)

5. Replace (1 – sin² θ) with cos² θ:
cos² θ / (cos² θ * sin θ)

6. Simplify:
1 / sin θ

7. Replace 1 / sin θ with cosec θ:
cosec θ

The correct answer should be cosec θ.

Correct Option: B

Ans: C

Explanation: Let ‘h’ be the height of the building. Let ‘x’ be the initial distance from the point on the ground to the building.

From the first scenario (angle of elevation = 45°):
tan(45°) = h/x
1 = h/x
x = h

From the second scenario (angle of elevation = 30°):
The distance is increased by 100 meters, so the new distance is x + 100
tan(30°) = h/(x + 100)
1/√3 = h/(x + 100)
x + 100 = h√3

Since x = h, substitute h for x:
h + 100 = h√3
100 = h√3 – h
100 = h(√3 – 1)
h = 100 / (√3 – 1)
Multiply the numerator and denominator by the conjugate (√3 + 1):
h = 100(√3 + 1) / ((√3 – 1)(√3 + 1))
h = 100(√3 + 1) / (3 – 1)
h = 100(√3 + 1) / 2
h = 50(√3 + 1)
h = 50√3 + 50