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JEE Mains Maths · Inverse Trigonometric Functions

Trigonometric Values of Inverse Expressions

Finding a sine, cosine or tangent of a sum or multiple of inverse functions by reading each angle off a right triangle and using the angle formulas.

All Inverse Trigonometric Functions notesJEE Mains Maths strategy

Why this matters

Eleven PYQs, ten of them multiple choice, and two from 2026. Six evaluate a sum of inverse functions or a trigonometric ratio of one; five apply the double, half or triple angle formulas to an inverse. Two ideas cover the page.

Concept 1 of 2: Reading ratios off a right triangle

Each inverse function names an angle. For A=tan⁡−1815A=\tan^{-1}\frac{8}{15}, draw a right triangle with opposite side 8 and adjacent side 15; the hypotenuse is 17, so sin⁡A=817\sin A=\frac8{17} and cos⁡A=1517\cos A=\frac{15}{17}. With every angle's sine and cosine known, the formulas for sin⁡(A±B)\sin(A\pm B) and cos⁡(A±B)\cos(A\pm B) finish the job.

Definition

  • sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B.
  • cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B.
  • tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}.
  • cos⁡−1\cos^{-1} of a negative number is obtuse: its sine is positive and its tangent negative.

Sum formula

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)=\sin A\cos B+\cos A\sin B

Worked example

Find sin⁡(tan⁡−1815+cos⁡−1725)\sin\left(\tan^{-1}\frac{8}{15}+\cos^{-1}\frac{7}{25}\right).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 Jan 2025 · Q60Moderate

Example 1 · Inverse Trigonometric Functions · Trigonometric Values of Inverse Expressions

cos⁡(sin⁡−135+sin⁡−1513+sin⁡−13365)\cos\left( \sin^{- 1}\frac{3}{5}+\sin^{- 1}\frac{5}{13}+\sin^{- 1}\frac{33}{65} \right) is equal to :

A triangle has no signs

A right triangle gives only positive ratios. For cos⁡−1\cos^{-1} of a negative number the angle is in (π2,π)\left(\frac\pi2,\pi\right), so its cosine and tangent are negative. For sin⁡−1\sin^{-1} or tan⁡−1\tan^{-1} of a negative number the angle is negative, so its sine and tangent are negative.
Drill 5 more on reading ratios off a right triangle

Concept 2 of 2: Double, half and triple angles of an inverse

If A=tan⁡−1tA=\tan^{-1}t, then tan⁡2A\tan2A, sin⁡2A\sin2A and cos⁡2A\cos2A are all rational in tt. So a multiple of an inverse function becomes an ordinary number. Convert every inverse to a tangent first, apply the multiple-angle formula, then combine with the sum formula for tangents.

Definition

  • With t=tan⁡At=\tan A: tan⁡2A=2t1−t2\tan2A=\frac{2t}{1-t^2}, sin⁡2A=2t1+t2\sin2A=\frac{2t}{1+t^2}, cos⁡2A=1−t21+t2\cos2A=\frac{1-t^2}{1+t^2}.
  • tan⁡3A=3t−t31−3t2\tan3A=\frac{3t-t^3}{1-3t^2}; cos⁡2A=1−2sin⁡2A\cos2A=1-2\sin^2A.
  • Half angle: tan⁡θ2=sin⁡θ1+cos⁡θ=1−cos⁡θsin⁡θ\tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta}.

Double angle in terms of the tangent

tan⁡2A=2t1−t2,sin⁡2A=2t1+t2,cos⁡2A=1−t21+t2\tan2A=\frac{2t}{1-t^2},\quad \sin2A=\frac{2t}{1+t^2},\quad \cos2A=\frac{1-t^2}{1+t^2}

Worked example

Find tan⁡(2tan⁡−112−tan⁡−117)\tan\left(2\tan^{-1}\frac12-\tan^{-1}\frac17\right).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q56Moderate

Example 2 · Inverse Trigonometric Functions · Trigonometric Values of Inverse Expressions

Considering the principal values of inverse trigonometric functions, the value of the expression tan⁡(2sin⁡−1(213)−2cos⁡−1(310))\tan\left( 2\sin^{- 1}\left( \frac{2}{\sqrt{13}} \right)- 2\cos^{- 1}\left( \frac{3}{\sqrt{10}} \right) \right) is equal to :

Pick the half angle that fits the range

Solving tan⁡θ=2t1−t2\tan\theta=\frac{2t}{1-t^2} for t=tan⁡θ2t=\tan\frac\theta2 gives two roots. When θ\theta is a principal value in (−π2,π2)\left(-\frac\pi2,\frac\pi2\right), θ2\frac\theta2 lies in (−π4,π4)\left(-\frac\pi4,\frac\pi4\right), so keep the root with ∣t∣<1|t|<1.
Drill 4 more on double, half and triple angles of an inverse

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 (2)

  • Reading ratios off a right triangle

    Sum formula

    sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)=\sin A\cos B+\cos A\sin B
  • Double, half and triple angles of an inverse

    Double angle in terms of the tangent

    tan⁡2A=2t1−t2,sin⁡2A=2t1+t2,cos⁡2A=1−t21+t2\tan2A=\frac{2t}{1-t^2},\quad \sin2A=\frac{2t}{1+t^2},\quad \cos2A=\frac{1-t^2}{1+t^2}

Watch out for (2)

Test yourself on Inverse Trigonometric Functions

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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