JEE Mains Maths · Trigonometric Identities

Compound Angle Formulae

The sine, cosine and tangent of a sum or difference of two angles, with the sign of each ratio fixed by its quadrant.

All Trigonometric Identities notes JEE Mains Maths strategy

Why this matters

Fourteen PYQs, thirteen of them multiple choice, and five from 2026. Three fix the sign of each ratio from the quadrant before anything else; six expand or spot sin(A ± B) and cos(A ± B), often to collapse a long expression into one ratio; five use tan(A ± B), two of them with tan A and tan B as the roots of a quadratic. Three ideas cover the page.

Concept 1 of 3: Signs by quadrant

One given ratio fixes the others up to sign. A right triangle gives their sizes:

\sin x=-\frac35

means sides 3, 4 and 5. The quadrant then gives each sign. When the question is about

\alpha+\beta

, place

\alpha+\beta

in its quadrant first.

Definition

Sizes come from the right triangle, or from $\sin^2x+\cos^2x=1$ .
Quadrant I: all ratios positive. II: only $\sin$ and $\csc$ . III: only $\tan$ and $\cot$ . IV: only $\cos$ and $\sec$ .
For $\alpha\pm\beta$ , add or subtract the ranges of $\alpha$ and $\beta$ , then use the sign of the given ratio to narrow the range.

Pythagorean identities

\sin^2x+\cos^2x=1,\quad1+\tan^2x=\sec^2x,\quad1+\cot^2x=\csc^2x

Worked example

\tan x=-\frac{5}{12}

and

\frac{3\pi}{2}<x<2\pi

. Find

13(\sin x-\cos x)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q73Moderate

Example 1 · Trigonometric Identities · Compound Angle Formulae

Let

\cos(\alpha + \beta) = - \frac{1}{10}

and

\sin(\alpha - \beta) = \frac{3}{8}

, where

0 < \alpha < \frac{\pi}{3}

and

0 < \beta < \frac{\pi}{4}

. If

\tan2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})},r,s \in \mathbb{N}

, then

r + s

is equal to

\_\_\_\_

The triangle gives the size, not the sign

\sin x=-\frac35

in quadrant III gives

\cos x=-\frac45

, not

+\frac45

. Fix every sign from the quadrant before substituting. For

\alpha+\beta

, place the sum, not the parts.

Drill 2 more on signs by quadrant

Concept 2 of 3: Sine and cosine of a sum

Expanding turns a condition such as

3\sin(\alpha+\beta)=2\sin(\alpha-\beta)

into a relation between

\tan\alpha

and

\tan\beta

. Read backwards, the same formulas collapse two products into one ratio. When a long expression has the shape

\sin P\cos Q\pm\cos P\sin Q

, regroup it until one angle is left.

Definition

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ .
$\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ .
Divide an expanded equation by $\cos A\cos B$ to get $\tan A$ and $\tan B$ .
$\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$ , $\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B$ .

Compound angles

\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,\quad\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B

Worked example

2\sin(\alpha-\beta)=\sin(\alpha+\beta)

. Find

\frac{\tan\alpha}{\tan\beta}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q60Moderate

Example 2 · Trigonometric Identities · Compound Angle Formulae

\cot x =\frac{5}{12}

for some

x \in\left( \pi,\frac{3\pi}{2} \right)

, then

\sin7x\left( \cos\frac{13x}{2}+ \sin\frac{13x}{2} \right)+ \cos7x\left( \cos\frac{13x}{2}- \sin\frac{13x}{2} \right)

is equal to

The sign in cos(A ± B) flips

\cos(A-B)=\cos A\cos B+\sin A\sin B

, with a plus. When collapsing a pair, match its sign to the formula, or

\cos(A-B)

becomes

\cos(A+B)

Drill 5 more on sine and cosine of a sum

Concept 3 of 3: Tangent of a sum

\tan(A+B)

needs only

\tan A+\tan B

and

\tan A\tan B

. Those are the sum and product of the roots when

\tan A

and

\tan B

solve a quadratic. When the sum of the angles is known, such as

45^\circ

, the same formula becomes a relation between the two tangents.

Definition

$\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$ .
If $\tan A,\tan B$ are the roots of $ax^2+bx+c=0$ : $\tan A+\tan B=-\frac ba$ , $\tan A\tan B=\frac ca$ .
$A+B=45^\circ$ gives $\tan A+\tan B+\tan A\tan B=1$ , so $(1+\tan A)(1+\tan B)=2$ .
$\tan(90^\circ-A)=\cot A$ , so $\tan A\tan(90^\circ-A)=1$ .

Tangent of a sum

\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}

Worked example

\tan A

and

\tan B

are the roots of

x^2-5x+6=0

, with

A,B\in\left(0,\frac{\pi}{2}\right)

. Find

A+B

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q56Moderate

Example 3 · Trigonometric Identities · Compound Angle Formulae

Let

\tan A,\tan B

, where

A,B \in\left( -\frac{\pi}{2},\frac{\pi}{2} \right)

, be the roots of the quadratic equation

x^{2}- 2x - 5 = 0

. Then

20\sin^{2}\left( \frac{A+B}{2} \right)

is equal to :

One tangent, two angles

\tan(A+B)=-1

fits both

\frac{3\pi}{4}

and

-\frac{\pi}{4}

. Use the ranges of

A

and

B

, and the signs of

\tan A

and

\tan B

, to choose the angle before taking its sine or cosine.

Drill 4 more on tangent of a sum

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)

Signs by quadrant
Pythagorean identities
$\sin^2x+\cos^2x=1,\quad1+\tan^2x=\sec^2x,\quad1+\cot^2x=\csc^2x$
Sine and cosine of a sum
Compound angles
$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,\quad\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$
Tangent of a sum
Tangent of a sum
$\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$

Watch out for (3)

The triangle gives the size, not the sign
→ Signs by quadrant
The sign in cos(A ± B) flips
→ Sine and cosine of a sum
One tangent, two angles
→ Tangent of a sum

Test yourself on Trigonometric Identities

15 past JEE Mains questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Trigonometric Identities chapter test Drill all 14 questions