Theory & Concepts

SSC CGL Trigonometric Ratios Questions, Formulas & Tricks

Get comprehensive theory, expert shortcuts, and hand-picked practice questions for Basic Trigonometric Ratios specifically designed for the SSC CGL 2025-26 pattern.

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25 min readDifficulty: Easy-Intermediate

Basic Trigonometry is entirely built around the properties of the right-angled triangle. By memorizing the core ratios, standard angle values, and complementary angle tricks, you can bypass long geometric proofs and solve Tier-1 questions visually in seconds.

Learning path

The Right-Angled Rules
Standard Angle Table
Complementary Angles
10 Exam-Level Examples

1. The Core Ratios

Every trigonometric function is simply a ratio between two sides of a right-angled triangle: Perpendicular (P), Base (B), and Hypotenuse (H).

Primary Ratios

$\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$

$\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}}$

$\tan\theta = \frac{\text{Perpendicular}}{\text{Base}}$

Mnemonic: SOH-CAH-TOA

Inverse Ratios

$\csc\theta = \frac{1}{\sin\theta} = \frac{H}{P}$

$\sec\theta = \frac{1}{\cos\theta} = \frac{H}{B}$

$\cot\theta = \frac{1}{\tan\theta} = \frac{B}{P}$

2. Standard Angle Values

You must memorize the values of these functions for standard angles (0°, 30°, 45°, 60°, 90°).

The Core Values

The three most commonly tested functions:

Angle	30°	45°	60°
$\sin\theta$	$1/2$	$1/\sqrt{2}$	$\sqrt{3}/2$
$\cos\theta$	$\sqrt{3}/2$	$1/\sqrt{2}$	$1/2$
$\tan\theta$	$1/\sqrt{3}$	$1$	$\sqrt{3}$

Notice that sine and cosine are mirrored. $\sin 30^\circ$ is equal to $\cos 60^\circ$ .

3. Complementary Angles Trick

If the sum of two angles is 90°, they are complementary. SSC heavily tests this to simplify ugly fractional expressions.

The Conversion Rule

$\sin(90^\circ - \theta) = \cos\theta$

$\tan(90^\circ - \theta) = \cot\theta$

$\sec(90^\circ - \theta) = \csc\theta$

Visual Shortcut

If you see $\frac{\sin 18^\circ}{\cos 72^\circ}$ , check the sum of the angles.

Since $18+72=90$ , they are exactly equal! The fraction simplifies instantly to 1.

4. 10 Solved examples

Question 01Exam Pattern

In a right triangle, if $ \sin\theta = 3/5 $, find the value of $ \cos\theta $.

4/5

5/4

3/4

5/3

Correct answer: a) 4/5

Solution

Step 1: We know $ \sin\theta = \frac{P}{H} $. So, $ P = 3 $ and $ H = 5 $.

Step 2: Use Pythagoras theorem: $ P^2 + B^2 = H^2 $.

Step 3: $ 3^2 + B^2 = 5^2 \implies 9 + B^2 = 25 \implies B^2 = 16 \implies B = 4 $.

Final calculation: $ \cos\theta = \frac{B}{H} = \frac{4}{5} $.

Question 02Exam Pattern

Find the exact value of $ \tan 45^\circ + \cos 60^\circ $.

1.5

2.5

Correct answer: b) 1.5

Solution

Step 1: Recall the standard angle values from the table.

Step 2: $ \tan 45^\circ = 1 $.

Step 3: $ \cos 60^\circ = \frac{1}{2} $.

Final calculation: $ 1 + \frac{1}{2} = 1.5 $.

Question 03Exam Pattern

If $ 3\tan\theta = 4 $, find the value of $ \frac{3\sin\theta + 2\cos\theta}{3\sin\theta - 2\cos\theta} $.

Correct answer: c) 3

Solution

Step 1: We are given $ \tan\theta = \frac{4}{3} $.

Step 2: Divide the numerator and denominator of the target expression by $ \cos\theta $.

Step 3: Expression becomes $ \frac{3\tan\theta + 2}{3\tan\theta - 2} $.

Step 4: Substitute $ \tan\theta = 4/3 $: $ \frac{3(4/3) + 2}{3(4/3) - 2} $.

Final calculation: $ \frac{4 + 2}{4 - 2} = \frac{6}{2} = 3 $.

Question 04Exam Pattern

Evaluate: $ \frac{\sin 18^\circ}{\cos 72^\circ} $.

Undefined

Correct answer: b) 1

Solution

Step 1: Notice that $ 18^\circ + 72^\circ = 90^\circ $. They are complementary angles.

Step 2: Use the rule $ \cos(90^\circ - \theta) = \sin\theta $.

Step 3: Let $ \theta = 18^\circ $. Then $ \cos(90^\circ - 18^\circ) = \cos 72^\circ = \sin 18^\circ $.

Final calculation: Since numerator and denominator are identical, $ \frac{\sin 18^\circ}{\sin 18^\circ} = 1 $.

Question 05Exam Pattern

If $ \tan(A+B) = \sqrt{3} $ and $ \tan(A-B) = \frac{1}{\sqrt{3}} $, find $ A $ and $ B $. (Given A > B)

A=30, B=30

A=45, B=15

A=60, B=30

A=15, B=45

Correct answer: b) A=45, B=15

Solution

Step 1: From standard tables, $ \tan 60^\circ = \sqrt{3} $. So, $ A + B = 60 $.

Step 2: Similarly, $ \tan 30^\circ = \frac{1}{\sqrt{3}} $. So, $ A - B = 30 $.

Step 3: Add both equations: $ (A + B) + (A - B) = 60 + 30 \implies 2A = 90 \implies A = 45 $.

Final calculation: Substitute $ A=45 $ into first eq: $ 45 + B = 60 \implies B = 15 $.

Question 06Exam Pattern

Find the value of $ \tan 1^\circ \times \tan 2^\circ \times \tan 3^\circ \dots \times \tan 89^\circ $.

Undefined

Correct answer: b) 1

Solution

Step 1: Pair up the complementary angles. $ \tan 1^\circ $ pairs with $ \tan 89^\circ $.

Step 2: We know $ \tan(90^\circ - \theta) = \cot\theta $. So, $ \tan 89^\circ = \cot 1^\circ $.

Step 3: The pair becomes $ \tan 1^\circ \times \cot 1^\circ $. Since $ \tan \times \cot = 1 $, the pair product is 1.

Step 4: This happens for all pairs (e.g. 2 and 88, 3 and 87).

Final calculation: The only term without a pair is $ \tan 45^\circ $, which equals 1. The total product is 1.

Question 07Exam Pattern

If $ \sin\theta = \cos(\theta - 30^\circ) $, find the value of $ \theta $.

30°

45°

60°

90°

Correct answer: c) 60°

Solution

Step 1: We know the complementary rule: $ \sin\theta = \cos(90^\circ - \theta) $.

Step 2: Substitute this into the equation: $ \cos(90^\circ - \theta) = \cos(\theta - 30^\circ) $.

Step 3: Since the cosines are equal, the angles must be equal: $ 90^\circ - \theta = \theta - 30^\circ $.

Step 4: Solve for $ \theta $: $ 2\theta = 120^\circ $.

Final calculation: $ \theta = 60^\circ $.

Question 08Exam Pattern

In a right triangle ABC, right-angled at B, AB = 24 cm, and BC = 7 cm. Determine $ \sin A $.

7/24

24/25

7/25

24/7

Correct answer: c) 7/25

Solution

Step 1: First, find the hypotenuse AC using Pythagoras theorem.

Step 2: $ AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625 $.

Step 3: $ AC = \sqrt{625} = 25 $ cm.

Step 4: $ \sin A = \frac{\text{Side opposite to A}}{\text{Hypotenuse}} = \frac{BC}{AC} $.

Final calculation: $ \sin A = \frac{7}{25} $.

Question 09Exam Pattern

If $ 5\cot\theta = 12 $, find the value of $ \csc\theta $.

13/5

12/13

5/13

13/12

Correct answer: a) 13/5

Solution

Step 1: Isolate the ratio: $ \cot\theta = \frac{12}{5} $.

Step 2: We know $ \cot\theta = \frac{B}{P} $. So, Base = 12, Perpendicular = 5.

Step 3: Use Pythagoras to find Hypotenuse: $ H = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 $.

Final calculation: $ \csc\theta = \frac{H}{P} = \frac{13}{5} $.

Question 010Exam Pattern

Evaluate: $ \sin^2 30^\circ + \cos^2 30^\circ $.

1/2

Correct answer: c) 1

Solution

Step 1: You can solve this two ways. Method 1 (Values): $ \sin 30^\circ = 1/2 $, $ \cos 30^\circ = \sqrt{3}/2 $.

Step 2: Square them: $ (1/2)^2 + (\sqrt{3}/2)^2 = 1/4 + 3/4 = 4/4 = 1 $.

Step 3: Method 2 (Identity): The universal trigonometric identity states $ \sin^2\theta + \cos^2\theta = 1 $ for any angle.

Final calculation: Result is 1.

SSC CGL Trigonometric Ratios Questions, Formulas & Tricks

Learning path

1. The Core Ratios

Primary Ratios

Inverse Ratios

2. Standard Angle Values

3. Complementary Angles Trick

The Conversion Rule

Visual Shortcut

4. 10 Solved examples

In a right triangle, if \( \sin\theta = 3/5 \), find the value of \( \cos\theta \).

Find the exact value of \( \tan 45^\circ + \cos 60^\circ \).

If \( 3\tan\theta = 4 \), find the value of \( \frac{3\sin\theta + 2\cos\theta}{3\sin\theta - 2\cos\theta} \).

Evaluate: \( \frac{\sin 18^\circ}{\cos 72^\circ} \).

If \( \tan(A+B) = \sqrt{3} \) and \( \tan(A-B) = \frac{1}{\sqrt{3}} \), find \( A \) and \( B \). (Given A > B)

Find the value of \( \tan 1^\circ \times \tan 2^\circ \times \tan 3^\circ \dots \times \tan 89^\circ \).

If \( \sin\theta = \cos(\theta - 30^\circ) \), find the value of \( \theta \).

In a right triangle ABC, right-angled at B, AB = 24 cm, and BC = 7 cm. Determine \( \sin A \).

If \( 5\cot\theta = 12 \), find the value of \( \csc\theta \).

Evaluate: \( \sin^2 30^\circ + \cos^2 30^\circ \).