Theory & Concepts

SSC CGL Trigonometric Identities Questions & Tricks

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

25 min readDifficulty: Hard

Trigonometric Identities form the hardest portion of Advanced Maths in SSC CGL Tier-2. However, 80% of these terrifying algebraic expressions can be collapsed in seconds using the "Inverse Hack" or by simply substituting standard angles to bypass the algebra entirely.

Learning path

  • The 3 Master Squares
  • The Sec-Tan Inverse Trick
  • Angle Value Putting
  • 10 Tier-2 Level Examples

1. The Core Square Identities

Every single complex manipulation stems from these three universal identities. They are non-negotiable and must be memorized perfectly.

Identity 1

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

Identity 2

sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1

Identity 3

csc⁡2θ−cot⁡2θ=1\csc^2\theta - \cot^2\theta = 1

2. The "Inverse Hack"

Because A2−B2=1A^2 - B^2 = 1 splits into (A+B)(A−B)=1(A+B)(A-B) = 1, we get a massive shortcut for the Sec-Tan and Csc-Cot pairs.

The Rule

If they give you the sum, the difference is exactly its reciprocal.

1. If sec⁡θ+tan⁡θ=X\sec\theta + \tan\theta = X, then sec⁡θ−tan⁡θ=1X\sec\theta - \tan\theta = \frac{1}{X}.

2. If csc⁡θ−cot⁡θ=Y\csc\theta - \cot\theta = Y, then csc⁡θ+cot⁡θ=1Y\csc\theta + \cot\theta = \frac{1}{Y}.

Once you have both equations, simply add or subtract them to isolate sec⁡\sec or tan⁡\tan.

3. Value Putting (The Bypass)

If an identity question has θ\theta in the problem but only numbers in the multiple-choice options, it means the answer is independent of the angle. Pick an angle and substitute!

Best Angles to Pick

For Sine/Cosine: Try 0∘0^\circ or 90∘90^\circ (Wipes out massive terms).

For Tangent/Cotangent: Try 45∘45^\circ (Turns them into 1).

The Danger Zone

NEVER pick an angle that makes a denominator 0, or results in an undefined function like tan⁡90∘\tan 90^\circ or csc⁡0∘\csc 0^\circ.

4. 10 Solved examples

Question 01Exam Pattern

If \( \sec\theta + \tan\theta = 3 \), find the value of \( \sec\theta \).

4/3
5/3
3/2
10/3
Correct answer: b) 5/3

Solution

Step 1: Use the Inverse Hack. Since \( \sec\theta + \tan\theta = 3 \), we instantly know \( \sec\theta - \tan\theta = 1/3 \).
Step 2: Add the two equations vertically to eliminate tangent.
Step 3: \( 2\sec\theta = 3 + 1/3 = 10/3 \).
Final calculation: Divide by 2. \( \sec\theta = 5/3 \).
Question 02Exam Pattern

Simplify: \( (1 + \tan^2\theta)(1 - \sin\theta)(1 + \sin\theta) \).

0
1
-1
2
Correct answer: b) 1

Solution

Step 1: The first bracket is an identity: \( 1 + \tan^2\theta = \sec^2\theta \).
Step 2: The second and third brackets form a difference of squares: \( (1 - \sin\theta)(1 + \sin\theta) = 1 - \sin^2\theta \).
Step 3: Replace \( 1 - \sin^2\theta \) with \( \cos^2\theta \).
Final calculation: \( \sec^2\theta \times \cos^2\theta = 1 \), because secant is the reciprocal of cosine.
Question 03Exam Pattern

Evaluate: \( \sin^6\theta + \cos^6\theta + 3\sin^2\theta\cos^2\theta \).

0
1
2
3
Correct answer: b) 1

Solution

Step 1: Use the Value Putting bypass since the options are constants.
Step 2: Substitute \( \theta = 0^\circ \).
Step 3: \( \sin 0^\circ = 0 \) and \( \cos 0^\circ = 1 \).
Final calculation: \( 0^6 + 1^6 + 3(0^2)(1^2) = 0 + 1 + 0 = 1 \).
Question 04Exam Pattern

Evaluate: \( (\sin^4\theta - \cos^4\theta + 1)\csc^2\theta \).

1
2
sin²θ
cos²θ
Correct answer: b) 2

Solution

Step 1: Factorize the first part as a difference of squares: \( (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta) \).
Step 2: Since \( \sin^2\theta + \cos^2\theta = 1 \), it simplifies to just \( \sin^2\theta - \cos^2\theta \).
Step 3: Add the 1: \( \sin^2\theta - \cos^2\theta + 1 \). Since \( 1 - \cos^2\theta = \sin^2\theta \), this becomes \( \sin^2\theta + \sin^2\theta = 2\sin^2\theta \).
Final calculation: Multiply by \( \csc^2\theta \): \( 2\sin^2\theta \times \csc^2\theta = 2(1) = 2 \).
Question 05Exam Pattern

If \( \csc\theta - \cot\theta = 1/2 \), find the value of \( \cos\theta \).

3/5
4/5
5/4
5/3
Correct answer: a) 3/5

Solution

Step 1: Inverse Hack. \( \csc\theta + \cot\theta = 2 \).
Step 2: Subtract the equations to isolate cotangent: \( 2\cot\theta = 2 - 1/2 = 3/2 \implies \cot\theta = 3/4 \).
Step 3: \( \cot\theta = B/P = 3/4 \). By Pythagoras, Hypotenuse \( H = 5 \).
Final calculation: \( \cos\theta = B/H = 3/5 \).
Question 06Exam Pattern

Simplify: \( \sqrt{\frac{1+\sin\theta}{1-\sin\theta}} \).

secθ - tanθ
secθ + tanθ
cscθ + cotθ
cscθ - cotθ
Correct answer: b) secθ + tanθ

Solution

Step 1: Rationalize the denominator by multiplying top and bottom by \( 1+\sin\theta \).
Step 2: Inside root: \( \frac{(1+\sin\theta)^2}{1-\sin^2\theta} \).
Step 3: Replace denominator with \( \cos^2\theta \). The root is now \( \sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}} \).
Step 4: Take the square root: \( \frac{1+\sin\theta}{\cos\theta} \).
Final calculation: Split the fraction: \( \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = \sec\theta + \tan\theta \).
Question 07Exam Pattern

Evaluate: \( \frac{1+\sin\theta}{\cos\theta} + \frac{\cos\theta}{1+\sin\theta} \).

2sinθ
2cosθ
2secθ
2cscθ
Correct answer: c) 2secθ

Solution

Step 1: Take the LCM of the denominators: \( \cos\theta(1+\sin\theta) \).
Step 2: Numerator becomes \( (1+\sin\theta)^2 + \cos^2\theta \).
Step 3: Expand: \( 1 + 2\sin\theta + \sin^2\theta + \cos^2\theta \).
Step 4: Combine identities: \( \sin^2\theta + \cos^2\theta = 1 \). Numerator is \( 2 + 2\sin\theta = 2(1+\sin\theta) \).
Final calculation: \( \frac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)} = \frac{2}{\cos\theta} = 2\sec\theta \).
Question 08Exam Pattern

If \( x = a\sin\theta \) and \( y = b\tan\theta \), find the value of \( \frac{a^2}{x^2} - \frac{b^2}{y^2} \).

0
1
a²+b²
-1
Correct answer: b) 1

Solution

Step 1: From the given equations, isolate the trig functions: \( \sin\theta = x/a \) and \( \tan\theta = y/b \).
Step 2: The target expression needs the inverse: \( \frac{a}{x} = \csc\theta \) and \( \frac{b}{y} = \cot\theta \).
Step 3: Square them: \( \frac{a^2}{x^2} = \csc^2\theta \) and \( \frac{b^2}{y^2} = \cot^2\theta \).
Final calculation: Substitute into expression: \( \csc^2\theta - \cot^2\theta = 1 \).
Question 09Exam Pattern

Evaluate: \( \sec^4\theta - \sec^2\theta \).

tan²θ + tan⁴θ
tan²θ - tan⁴θ
1 - tan⁴θ
1 + tan⁴θ
Correct answer: a) tan²θ + tan⁴θ

Solution

Step 1: Factor out the common term: \( \sec^2\theta(\sec^2\theta - 1) \).
Step 2: Apply the identity \( \sec^2\theta - 1 = \tan^2\theta \). Expression becomes \( \sec^2\theta \times \tan^2\theta \).
Step 3: Convert the remaining secant into tangent: \( (1 + \tan^2\theta)\tan^2\theta \).
Final calculation: Expand the bracket: \( \tan^2\theta + \tan^4\theta \).
Question 010Exam Pattern

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

√21
±√21
5
±5
Correct answer: b) ±√21

Solution

Step 1: Use the SSC Master formula for this pattern: If \( a\cos\theta - b\sin\theta = c \) and \( a\sin\theta + b\cos\theta = x \).
Step 2: Squaring and adding both equations yields the identity: \( a^2 + b^2 = c^2 + x^2 \).
Step 3: Substitute the known values: \( 3^2 + 4^2 = 2^2 + x^2 \).
Step 4: \( 9 + 16 = 4 + x^2 \implies 25 = 4 + x^2 \implies x^2 = 21 \).
Final calculation: \( x = \pm\sqrt{21} \).