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SSC CGL Maths · Trigonometric Ratios and Identities

Simplifying and Proving Identities

Three Pythagorean identities, used to simplify an expression to one ratio or a number, to spot a false identity, and to remove θ from a pair of equations.

All Trigonometric Ratios and Identities notes

Why this matters

49 questions in the bank, and 18 of them are MODERATE, more than on any earlier page. Nearly every one ends in sin²θ + cos²θ = 1 or one of its two cousins. When the options are expressions in θ, putting in one angle settles most of them faster than the algebra.

Concept 1 of 3: The three Pythagorean identities

Divide Pythagoras, opp2+adj2=hyp2\text{opp}^2 + \text{adj}^2 = \text{hyp}^2, by hyp2\text{hyp}^2, by adj2\text{adj}^2 or by opp2\text{opp}^2, and you get the three identities. They hold for every angle, which is what makes them identities.

Definition

  • sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, so 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta and 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta.
  • 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta, so sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1 and 11+tan⁡2θ=cos⁡2θ\frac{1}{1 + \tan^2\theta} = \cos^2\theta.
  • 1+cot⁡2θ=cosec2θ1 + \cot^2\theta = \text{cosec}^2\theta, so cosec2θ−cot⁡2θ=1\text{cosec}^2\theta - \cot^2\theta = 1 and 11+cot⁡2θ=sin⁡2θ\frac{1}{1 + \cot^2\theta} = \sin^2\theta.
  • Spotting a false identity, fast: put θ = 45° (or 0°) into each statement. One angle where it fails proves it false. One angle where it holds proves nothing, so use this only to reject.
  • Square roots: 1−sin⁡2θ=cos⁡θ\sqrt{1 - \sin^2\theta} = \cos\theta only when cos⁡θ≥0\cos\theta \ge 0, which is true for every acute angle.

Pythagorean identities

sin⁡2θ+cos⁡2θ=11+tan⁡2θ=sec⁡2θ1+cot⁡2θ=cosec2θ\sin^2\theta + \cos^2\theta = 1 \qquad 1 + \tan^2\theta = \sec^2\theta \qquad 1 + \cot^2\theta = \text{cosec}^2\theta

Worked example

Simplify (sec⁡2θ−1)(1−sin⁡2θ)(\sec^2\theta - 1)(1 - \sin^2\theta).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 21 Sep 2025, Shift 2 · Q67Easy

Example 1 · Trigonometric Ratios and Identities · Simplifying and Proving Identities

If cot⁡A=x+1x\cot A = x + \frac{1}{x}, then what is cosec2A\text{cosec}^2 A?

A minus sign that is not an identity

cos⁡2θ−sin⁡2θ=1\cos^2\theta - \sin^2\theta = 1 is false: at 45° the left side is 0. Only the sum of the squares of sin and cos is 1.

Moving 1 across with the wrong sign

From 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta, tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1, never 1−sec⁡2θ1 - \sec^2\theta. The second would make tan⁡2θ\tan^2\theta negative for every angle.

One passing angle is not a proof

A false statement can still hold at one angle: sin⁡θ=cos⁡θ\sin\theta = \cos\theta is true at 45°. Use a test angle to throw options out, never to accept one on its own.
Drill 13 more on the three pythagorean identities

Concept 2 of 3: Rewrite in sin and cos, then simplify

Most long expressions shrink once everything is written in sin and cos. Add the fractions, and somewhere a 1−sin⁡2θ1 - \sin^2\theta or a 1−cos⁡2θ1 - \cos^2\theta appears and collapses into a single square.

Definition

The full method, in order:

  • Replace tan, cot, sec and cosec by sin and cos.
  • Put fractions over a common denominator. With 1±sin⁡θ1 \pm \sin\theta or 1±cos⁡θ1 \pm \cos\theta below, the product is cos⁡2θ\cos^2\theta or sin⁡2θ\sin^2\theta.
  • Factor: a2−b2a^2 - b^2, a3+b3a^3 + b^3, a3−b3a^3 - b^3. With a=sin⁡θa = \sin\theta and b=cos⁡θb = \cos\theta, a2+b2=1a^2 + b^2 = 1 inside the bracket.
  • Cancel only common factors, never terms joined by + or −.

The fast route for options in θ: put a convenient angle into the question and into each option, and keep the option that matches.

  • Use 45°, 60° or 30°. Avoid 90° when tan or sec appears, and 0° when cot or cosec appears.
  • If two options give the same value at your angle, try a second angle.

Factor patterns with a² + b² = 1

a3+b3=(a+b)(a2−ab+b2)a3−b3=(a−b)(a2+ab+b2)(1−cos⁡θ)(1+cos⁡θ)=sin⁡2θa^3 + b^3 = (a + b)(a^2 - ab + b^2) \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2) \qquad (1 - \cos\theta)(1 + \cos\theta) = \sin^2\theta

Worked example

Simplify cos⁡θ1−sin⁡θ−cos⁡θ1+sin⁡θ\frac{\cos\theta}{1 - \sin\theta} - \frac{\cos\theta}{1 + \sin\theta}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 18 Sep 2024, Shift 2 · Q58Moderate

Example 2 · Trigonometric Ratios and Identities · Simplifying and Proving Identities

Simplify. sin⁡A1+cos⁡A+cot⁡A\frac{\sin A}{1 + \cos A} + \cot A, where A is an acute angle.

Cancelling across a plus sign

In sin⁡θ+cos⁡θsin⁡θ\frac{\sin\theta + \cos\theta}{\sin\theta} the sin does not cancel; the fraction is 1+cot⁡θ1 + \cot\theta. Split the fraction or factor first, then cancel.

A test angle where two options agree

At θ = 90°, sin⁡θ1+cos⁡θ\frac{\sin\theta}{1 + \cos\theta} and sin⁡θ1−cos⁡θ\frac{\sin\theta}{1 - \cos\theta} are both 1, yet they are different expressions. If two options tie at your angle, test again at 60°.
Drill 20 more on rewrite in sin and cos, then simplify

Concept 3 of 3: Eliminating θ from two equations

When x and y are both built from the same θ, look for the identity whose two pieces are x and y. Squaring and then adding or subtracting removes θ and leaves a relation in x and y alone.

Definition

  • x=acos⁡θx = a\cos\theta, y=bsin⁡θy = b\sin\theta: x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.
  • x=asec⁡θx = a\sec\theta, y=btan⁡θy = b\tan\theta: x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. The same holds for cosec and cot.
  • (sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2=2(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2, and more generally the two expressions acos⁡θ+bsin⁡θa\cos\theta + b\sin\theta and asin⁡θ−bcos⁡θa\sin\theta - b\cos\theta have squares that add to a2+b2a^2 + b^2. The cross terms cancel.
  • Two equations with the same left part: divide one by the other, and the shared part goes, leaving a tan.
  • Fast route: the relation is true for every θ, so put θ = 0° (or 60°) into both equations and into each option.
  • Limit: choose an angle that keeps every quantity defined. sec 90° and cosec 0° do not exist.

Removing θ

x=acos⁡θ, y=bsin⁡θ⇒x2a2+y2b2=1x=asec⁡θ, y=btan⁡θ⇒x2a2−y2b2=1(acos⁡θ+bsin⁡θ)2+(asin⁡θ−bcos⁡θ)2=a2+b2x = a\cos\theta,\ y = b\sin\theta \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad x = a\sec\theta,\ y = b\tan\theta \Rightarrow \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \qquad (a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 = a^2 + b^2

Worked example

If x=5 cosec θx = 5\,\text{cosec}\,\theta and y=5cot⁡θy = 5\cot\theta, find x2−y2x^2 - y^2.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 25 Sep 2024, Shift 2 · Q72Moderate

Example 3 · Trigonometric Ratios and Identities · Simplifying and Proving Identities

If x=psec⁡Acos⁡Bx = p \sec A \cos B, y=qsec⁡Asin⁡By = q \sec A \sin B and z=rtan⁡Az = r \tan A, what is the value of the following expression? x2p2+y2q2−z2r2\frac{x^2}{p^2} + \frac{y^2}{q^2} - \frac{z^2}{r^2}

Adding where the identity subtracts

sec and tan go together with a minus: sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1. Only sin and cos add their squares to 1. Pair the two given quantities with the right identity before squaring.

Expanding without checking the cross terms

(acos⁡θ+bsin⁡θ)2(a\cos\theta + b\sin\theta)^2 has a middle term 2absin⁡θcos⁡θ2ab\sin\theta\cos\theta. It cancels only if the other expression carries the same term with the opposite sign, so write both middle terms down.
Drill 13 more on eliminating θ from two equations

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • The three Pythagorean identities

    Pythagorean identities

    sin⁡2θ+cos⁡2θ=11+tan⁡2θ=sec⁡2θ1+cot⁡2θ=cosec2θ\sin^2\theta + \cos^2\theta = 1 \qquad 1 + \tan^2\theta = \sec^2\theta \qquad 1 + \cot^2\theta = \text{cosec}^2\theta
  • Rewrite in sin and cos, then simplify

    Factor patterns with a² + b² = 1

    a3+b3=(a+b)(a2−ab+b2)a3−b3=(a−b)(a2+ab+b2)(1−cos⁡θ)(1+cos⁡θ)=sin⁡2θa^3 + b^3 = (a + b)(a^2 - ab + b^2) \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2) \qquad (1 - \cos\theta)(1 + \cos\theta) = \sin^2\theta
  • Eliminating θ from two equations

    Removing θ

    x=acos⁡θ, y=bsin⁡θ⇒x2a2+y2b2=1x=asec⁡θ, y=btan⁡θ⇒x2a2−y2b2=1(acos⁡θ+bsin⁡θ)2+(asin⁡θ−bcos⁡θ)2=a2+b2x = a\cos\theta,\ y = b\sin\theta \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad x = a\sec\theta,\ y = b\tan\theta \Rightarrow \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \qquad (a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 = a^2 + b^2

Watch out for (7)

Test yourself on Trigonometric Ratios and Identities

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