Trigonometry: Aptitude Questions with Answers – Free Practice!

Ans: A

Explanation: We can rewrite the expression using the property that cot(x) = tan(90 – x).

Given expression is: tan(35°)cot(40°)tan(45°)cot(50°)tan(55°)

We know tan(45°) = 1. So, the expression becomes: tan(35°)cot(40°)cot(50°)tan(55°)

We can rewrite cot(40°) as tan(90°-40°) = tan(50°) and cot(50°) as tan(90°-50°) = tan(40°). But since we have both tan(50°) and cot(50°), and tan(55°), we can rewrite it like this:

cot(40°) = tan(90° – 40°) = tan(50°)
cot(50°) = tan(90° – 50°) = tan(40°)

Now, the expression becomes: tan(35°) * tan(50°) * 1 * tan(40°) * tan(55°)

Let’s pair the terms:
tan(35°) and tan(55°) = tan(35°) * tan(90° – 35°) = tan(35°) * cot(35°) = 1
tan(40°) and tan(50°) = tan(40°) * tan(90° – 40°) = tan(40°) * cot(40°) = 1

Therefore the expression becomes 1 * 1 * 1 = 1

Ans: D

Explanation: We need to evaluate the given trigonometric expression. Recall the following values:
* tan 30° = 1/√3
* sec 30° = 2/√3
* csc 45° = √2
* cot 45° = 1
* sec 60° = 2
* csc 60° = 2/√3

Now substitute these values into the expression:
\(\frac{\tan^2 30^\circ}{\sec^2 30^\circ} + \frac{\csc^2 45^\circ}{\cot^2 45^\circ} – \frac{\sec^2 60^\circ}{\csc^2 60^\circ}\) = \(\frac{(1/\sqrt{3})^2}{(2/\sqrt{3})^2} + \frac{(\sqrt{2})^2}{1^2} – \frac{2^2}{(2/\sqrt{3})^2}\)
= \(\frac{1/3}{4/3} + \frac{2}{1} – \frac{4}{4/3}\)
= \(\frac{1}{4} + 2 – 3\)
= \(\frac{1}{4} – 1\)
= \(\frac{1-4}{4}\)
= \(-\frac{3}{4}\)

Ans: C

Explanation:
First, we simplify the numerator. We have tan 45° = 1. We can also use the identities tan(90° – x) = cot x and cot x = 1/tan x.
tan 77° = tan(90° – 13°) = cot 13° = 1/tan 13°
tan 54° = tan(90° – 36°) = cot 36° = 1/tan 36°
So, the numerator becomes:
tan 13° * tan 36° * 1 * (1/tan 36°) * (1/tan 13°) = 1

Next, we simplify the denominator:
sec 60° = 2, so sec² 60° = 4.
sin 60° = √3/2, so sin² 60° = (√3/2)² = 3/4.
cos 60° = 1/2
The denominator becomes:
2 * 4 * (3/4 – 3 * (1/2) + 2)
= 8 * (3/4 – 3/2 + 2)
= 8 * (3/4 – 6/4 + 8/4)
= 8 * (5/4)
= 10

Therefore, the expression simplifies to 1/10.

Ans: C

Explanation: First, we evaluate each trigonometric term:
* sin 30° = 1/2, so sin² 30° = (1/2)² = 1/4
* tan 60° = √3
* cos 60° = 1/2, so cos² 60° = (1/2)² = 1/4
* sec 30° = 2/√3, so sec² 30° = (2/√3)² = 4/3
* cot 45° = 1, so cot² 45° = 1² = 1
* sec 60° = 2, so sec² 60° = 2² = 4
* sin 60° = √3/2, so sin² 60° = (√3/2)² = 3/4
* cos 90° = 0, so cos² 90° = 0² = 0

Now, substitute these values into the expression:

Numerator: 2 * (1/4) * √3 – 3 * (1/4) * (4/3) = (√3/2) – 1
Denominator: 4 * 1 – 4 + 3/4 + 0 = 3/4

So the expression becomes:
((√3/2) – 1) / (3/4) = (√3 – 2)/2 * 4/3 = 2(√3 – 2)/3

Ans: A

Explanation: The string, the height of the kite, and the ground form a right-angled triangle. The string is the hypotenuse (420m), the height is the opposite side to the 30° angle, and the ground is the adjacent side. We use the sine function: sin(angle) = opposite/hypotenuse. sin(30°) = height/420. We know sin(30°) = 0.5 or 1/2. Therefore, height = 420 * 0.5 = 210 m.

Ans: B

Explanation:
Let the trees be A and B, and the point on the road be P. Let the height of tree A be 200 meters (angle of depression 45 degrees) and the height of tree B be h meters (angle of depression 60 degrees). Let the distance of point P from the base of tree A be x meters, and the distance from the base of tree B be (400 – x) meters.
For tree A: tan(45°) = (height of A) / x => 1 = 200 / x => x = 200 meters
For tree B: tan(60°) = h / (400 – x) => √3 = h / (400 – 200) => √3 = h / 200 => h = 200√3 meters.

Ans: C

Explanation: Let the height of the tree be h = 45 m, and the length of the shadow be x = \(15\sqrt{3}\) m. The angle of elevation of the sun, θ, can be found using the tangent function: tan(θ) = (opposite side) / (adjacent side) = h / x. So, tan(θ) = 45 / \(15\sqrt{3}\) = 3 / \(\sqrt{3}\) = \(\sqrt{3}\). We know that tan(60°) = \(\sqrt{3}\). Therefore, the angle of elevation of the sun is 60°.

Ans: C

Explanation: Let the height of the tower be ‘h’ meters. Let the initial length of the shadow be x.

When the sun’s altitude is 45°, tan(45°) = h/x => 1 = h/x => x = h
When the sun’s altitude is 30°, the shadow’s length is x + 8.4. So, tan(30°) = h/(x+8.4) => 1/√3 = h/(h+8.4) [Since x=h]

Therefore, h + 8.4 = h√3
8.4 = h√3 – h
8.4 = h(√3 – 1)
h = 8.4 / (√3 – 1)
Multiplying numerator and denominator by (√3 + 1),
h = 8.4(√3 + 1) / (3 – 1)
h = 8.4(√3 + 1) / 2
h = 4.2(√3 + 1)

Ans: D

Explanation: Let the height of the building be ‘h’ meters, the height of the tower be T=72 meters, the angle of elevation from the top of the building to the top of the tower be 30 degrees, and the angle of elevation from the bottom of the building to the top of the tower be 60 degrees.

Let the distance between the building and the tower be ‘x’.
From the bottom of the building: tan(60) = 72/x => x = 72/sqrt(3) = 24*sqrt(3) meters.
From the top of the building: the vertical distance from the top of the building to the top of the tower is 72 – h.
tan(30) = (72-h) / x
1/sqrt(3) = (72-h) / (24*sqrt(3))
24*sqrt(3) / sqrt(3) = 72 – h
24 = 72 – h
h = 72 – 24
h = 48 meters

Ans: D

Explanation: We need to evaluate the expression \(\cot 45^\circ – \frac{1}{\sqrt{3}}\csc 60^\circ\).

We know that:
– \(\cot 45^\circ = 1\)
– \(\csc 60^\circ = \frac{2}{\sqrt{3}}\)

Substituting these values into the expression, we get:
\(1 – \frac{1}{\sqrt{3}} \cdot \frac{2}{\sqrt{3}} = 1 – \frac{2}{3} = \frac{3}{3} – \frac{2}{3} = \frac{1}{3}\)

Ans: D

Explanation: We need to evaluate the given trigonometric expression.
First, let’s find the values of the trigonometric functions:
* tan 45° = 1
* sec 60° = 2
* sin 30° = 1/2
* cos 90° = 0
* sin 60° = √3/2
* csc 90° = 1
* cot 90° = 0

Now, substitute these values into the expression:
Numerator: (tan² 45° + sec² 60° * sin² 30° + cos² 90° * sin² 60°)
= (1² + 2² * (1/2)² + 0² * (√3/2)²)
= (1 + 4 * (1/4) + 0)
= (1 + 1 + 0)
= 2

Denominator: (csc² 90° + cot 90° * sin 60°)
= (1² + 0 * (√3/2))
= (1 + 0)
= 1

So the expression is 2 / 1 = 2.

Ans: A

Explanation: We can simplify the given expression step-by-step.

First, let’s simplify the numerator using the double angle identity:

1 – 2sin²θcos²θ = 1 – (2sinθcosθ)² = 1 – (sin2θ)² = 1 – sin²(2θ)

Next, let’s simplify the denominator:

sin⁴θ + cos⁴θ = (sin²θ + cos²θ)² – 2sin²θcos²θ = 1 – 2sin²θcos²θ

Now, substitute these simplified expressions back into the original expression:

(1 – 2sin²θcos²θ) / (sin⁴θ + cos⁴θ) – 1 = (1 – 2sin²θcos²θ) / (1 – 2sin²θcos²θ) – 1
= 1 – 1
= 0

Therefore, the value of the expression is 0.

Ans: C

Explanation:
Given that \(\sin A = \frac{1}{2}\) and A is an acute angle. Since \(\sin 30^\circ = \frac{1}{2}\), we have \(A = 30^\circ\).
Now, we calculate the trigonometric values needed:
\(\tan A = \tan 30^\circ = \frac{1}{\sqrt{3}}\)
\(\cot A = \cot 30^\circ = \sqrt{3}\)
\(\csc A = \csc 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{\frac{1}{2}} = 2\)
Substitute these values into the expression:
\(\frac{\tan A – \cot A}{\sqrt{3}(1 + \csc A)} = \frac{\frac{1}{\sqrt{3}} – \sqrt{3}}{\sqrt{3}(1 + 2)} = \frac{\frac{1 – 3}{\sqrt{3}}}{3\sqrt{3}} = \frac{\frac{-2}{\sqrt{3}}}{3\sqrt{3}} = \frac{-2}{\sqrt{3} \cdot 3\sqrt{3}} = \frac{-2}{3 \cdot 3} = \frac{-2}{9}\)

Ans: A

Explanation: Given that cos(θ) = 35/37. We can use the Pythagorean identity sin²(θ) + cos²(θ) = 1 to find sin(θ).
sin²(θ) = 1 – cos²(θ) = 1 – (35/37)² = 1 – 1225/1369 = (1369 – 1225)/1369 = 144/1369
Therefore, sin(θ) = √(144/1369) = 12/37.
Now, csc(θ) = 1/sin(θ) = 1/(12/37) = 37/12.

Ans: B

Explanation: We can simplify the expression using trigonometric identities.

1. sin(90° – θ) = cos θ
2. cos(π – θ) = -cos θ
3. cos(π/2 – θ) = sin θ
4. sin(π – θ) = sin θ

Applying these identities:

Numerator: sin(90° – 10θ) – cos(π – 6θ) = cos(10θ) – (-cos(6θ)) = cos(10θ) + cos(6θ)
Denominator: cos(π/2 – 10θ) – sin(π – 6θ) = sin(10θ) – sin(6θ)

Now, use the sum-to-product identities:

* cos A + cos B = 2cos((A+B)/2)cos((A-B)/2)
* sin A – sin B = 2cos((A+B)/2)sin((A-B)/2)

Applying these:

Numerator: cos(10θ) + cos(6θ) = 2cos((10θ + 6θ)/2)cos((10θ – 6θ)/2) = 2cos(8θ)cos(2θ)
Denominator: sin(10θ) – sin(6θ) = 2cos((10θ + 6θ)/2)sin((10θ – 6θ)/2) = 2cos(8θ)sin(2θ)

The original expression becomes:

(2cos(8θ)cos(2θ)) / (2cos(8θ)sin(2θ))

The 2cos(8θ) terms cancel, leaving:

cos(2θ) / sin(2θ) = cot(2θ)

Ans: A

Explanation: We can use trigonometric identities to solve this. First, we know that \(\sin^2 x + \sin^2 (90^\circ – x) = \sin^2 x + \cos^2 x = 1\). Since \(42^\circ = 90^\circ – 48^\circ\), we have \(\sin^2 48^\circ + \sin^2 42^\circ = 1\). Next, we know that \(\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}\), and therefore, \(\sec^2 30^\circ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}\). Also, we know that \(\tan 60^\circ = \sqrt{3}\), and therefore, \(\tan^2 60^\circ = (\sqrt{3})^2 = 3\). Substituting these values into the expression, we have \(1 – \frac{4}{3} + 3 = 4 – \frac{4}{3} = \frac{12 – 4}{3} = \frac{8}{3}\).

Ans: D

Explanation:
First, expand the given equation:
6tan²A + 6tanA = 5 – tanA
Rearrange to form a quadratic equation in tanA:
6tan²A + 7tanA – 5 = 0
Now, solve the quadratic equation for tanA. We can factorize the equation:
(3tanA + 5)(2tanA – 1) = 0
This gives us two possible solutions for tanA:
tanA = -5/3 or tanA = 1/2
Since 0 < A < π/2, tanA must be positive. Therefore, tanA = 1/2.
Now, we can find sinA and cosA using tanA = 1/2. We can imagine a right-angled triangle where the opposite side is 1 and the adjacent side is 2. Using the Pythagorean theorem, the hypotenuse will be √(1² + 2²) = √5.
Thus, sinA = 1/√5 and cosA = 2/√5.
Now, calculate sinA + cosA:
sinA + cosA = 1/√5 + 2/√5 = 3/√5
The answer is 3/√5, which can also be written as 3/√5 = (3√5)/5 if you rationalize the denominator. However, none of the given options matches (3√5)/5. Instead, we can multiply numerator and denominator of 3/√5 by sqrt(5) to get 3/sqrt(5).

Ans: A

Explanation: We are given two equations:
1) `x sin^3 θ + y cos^3 θ = sin θ cos θ`
2) `x sin θ – y cos θ = 0`

From equation (2), we can express `x` in terms of `y`:
`x sin θ = y cos θ`
`x = y (cos θ / sin θ)`
`x = y cot θ`

Now, substitute the value of `x` in equation (1):
`y cot θ sin^3 θ + y cos^3 θ = sin θ cos θ`
`y (cos θ / sin θ) sin^3 θ + y cos^3 θ = sin θ cos θ`
`y cos θ sin^2 θ + y cos^3 θ = sin θ cos θ`
`y cos θ (sin^2 θ + cos^2 θ) = sin θ cos θ`
Since `sin^2 θ + cos^2 θ = 1`,
`y cos θ = sin θ cos θ`
If `cos θ ≠ 0`, then `y = sin θ`
Substitute y = sin θ in equation (2), `x sin θ – sin θ cos θ = 0`
`x sin θ = sin θ cos θ`
If `sin θ ≠ 0`, then `x = cos θ`

Now, we can find `x^2 + y^2`:
`x^2 + y^2 = cos^2 θ + sin^2 θ = 1`

Ans: C

Explanation:
Given that \(\sin(3\theta)\sec(2\theta) = 1\), which implies \(\sin(3\theta) = \cos(2\theta)\).
We know that \(\sin(x) = \cos(90^\circ – x)\). Therefore,
\(3\theta = 90^\circ – 2\theta\)
\(5\theta = 90^\circ\)
\(\theta = 18^\circ\)
Now we need to find the value of \(3\tan^2(5\theta/2) – 1\).
\(\frac{5\theta}{2} = \frac{5 \cdot 18^\circ}{2} = \frac{90^\circ}{2} = 45^\circ\)
\(\tan(45^\circ) = 1\)
So, \(3\tan^2(5\theta/2) – 1 = 3\tan^2(45^\circ) – 1 = 3(1)^2 – 1 = 3 – 1 = 2\)

Ans: D

Explanation: Let’s expand the expression using the trigonometric identity: sin(x)cos(y) = 0 if x = y.
sin(B-C)cos(A-D) + sin(A-B)cos(C-D) + sin(C-A)cos(B-D)
Consider the case where A=B=C=D. Then, each term in the sum becomes:
sin(0)cos(0) + sin(0)cos(0) + sin(0)cos(0) = 0 + 0 + 0 = 0

We can also solve this by setting A=B=C=D.