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Inverse Trigonometric Functions formulas

13 formulas and 13 common traps for JEE Mains Maths Inverse Trigonometric Functions, grouped by subtopic.

Full notes with worked examples

Domain, Range and Principal Values

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Domain conditions on the argument

Domains

sin⁡−1u, cos⁡−1u: ∣u∣≤1,sec⁡−1u, csc⁡−1u: ∣u∣≥1\sin^{-1}u,\ \cos^{-1}u:\ |u|\le1,\qquad \sec^{-1}u,\ \csc^{-1}u:\ |u|\ge1

Ranges and complementary pairs

Complementary pairs

sin⁡−1x+cos⁡−1x=tan⁡−1x+cot⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}

Inverse of a trigonometric value outside the principal range

Back into the principal range

sin⁡−1(sin⁡x)=π−x  (π2≤x≤3π2),cos⁡−1(cos⁡x)=2π−x  (π≤x≤2π)\sin^{-1}(\sin x)=\pi-x\ \ \left(\tfrac{\pi}{2}\le x\le\tfrac{3\pi}{2}\right),\qquad \cos^{-1}(\cos x)=2\pi-x\ \ (\pi\le x\le2\pi)

Common traps

Do not multiply by a denominator of unknown sign

Turning pq≤1\frac{p}{q}\le1 into p≤qp\le q is wrong when q<0q<0. Square instead: ∣pq∣≤1\left|\frac pq\right|\le1 exactly when p2≤q2p^2\le q^2, and then remove the points where q=0q=0.

Check whether each endpoint is reached

tan⁡−1x\tan^{-1}x never equals ±π2\pm\frac\pi2, and x2x2+1\frac{x^2}{x^2+1} never equals 1. An endpoint that is only approached gets a round bracket, and options often differ only in that bracket.

sin⁻¹(sin x) is not always x

sin⁡−1(sin⁡3)\sin^{-1}(\sin3) is π−3\pi-3, not 3, because 3 lies outside [−π2,π2]\left[-\frac\pi2,\frac\pi2\right]. Before writing the answer, check that it lies in the principal range.

Trigonometric Values of Inverse Expressions

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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}

Common traps

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.

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.

Simplifying Inverse Functions of a Variable

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Substitute and track the interval

Standard substitutions

2tan⁡−1x=sin⁡−12x1+x2  (∣x∣≤1),2tan⁡−1x=cos⁡−11−x21+x2  (x≥0)2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}\ \ (|x|\le1),\qquad 2\tan^{-1}x=\cos^{-1}\frac{1-x^2}{1+x^2}\ \ (x\ge0)

Turning a composition into algebra

Composition to algebra

sin⁡(tan⁡−1x)=x1+x2,cos⁡(sin⁡−1x)=1−x2\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^2}},\qquad \cos(\sin^{-1}x)=\sqrt{1-x^2}

Complementary identities with a condition

Complementary identity

sin⁡−1x+cos⁡−1x=π2,−1≤x≤1\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\quad -1\le x\le1

Common traps

The interval for x decides the formula

sin⁡−1(2x1−x2)\sin^{-1}\left(2x\sqrt{1-x^2}\right) is 2sin⁡−1x2\sin^{-1}x only for ∣x∣≤12|x|\le\frac1{\sqrt2}. Outside that interval it is π−2sin⁡−1x\pi-2\sin^{-1}x or −π−2sin⁡−1x-\pi-2\sin^{-1}x. Read the interval before using a standard result.

Squaring brings extra roots

cos⁡(sin⁡−1x)=1−x2\cos(\sin^{-1}x)=\sqrt{1-x^2} is never negative, but tan⁡(cos⁡−1x)=1−x2x\tan(\cos^{-1}x)=\frac{\sqrt{1-x^2}}x takes the sign of xx. After squaring an equation, put each root back and check its sign.

The identity needs the same argument

sin⁡−1x+cos⁡−1y=π2\sin^{-1}x+\cos^{-1}y=\frac\pi2 holds only when x=yx=y. And sec⁡−1x+csc⁡−1x=π2\sec^{-1}x+\csc^{-1}x=\frac\pi2 needs ∣x∣≥1|x|\ge1. Check the arguments match before replacing a sum by π2\frac\pi2.

Sums of Inverse Tangents and Telescoping Series

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The addition formula for inverse tangents

Addition formula

tan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab(ab<1)\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}\quad(ab<1)

Telescoping sums of inverse tangents

Telescoping term

tan⁡−1b−a1+ab=tan⁡−1b−tan⁡−1a\tan^{-1}\frac{b-a}{1+ab}=\tan^{-1}b-\tan^{-1}a

Common traps

When the product exceeds 1

For positive a,ba,b with ab>1ab>1, a+b1−ab\frac{a+b}{1-ab} is negative although both angles are positive. The formula alone gives a negative angle; add π\pi.

Keep the order of the difference

The term must be tan⁡−1(next)−tan⁡−1(this)\tan^{-1}(\text{next})-\tan^{-1}(\text{this}), so the numerator is the larger minus the smaller. With the order reversed the sum comes out with the wrong sign, and it will not match the options.

Equations in Inverse Trigonometric Functions

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Equations in inverse tangents

Taking tangents

tan⁡−1a+tan⁡−1b=c ⇒ a+b1−ab=tan⁡c\tan^{-1}a+\tan^{-1}b=c\ \Rightarrow\ \frac{a+b}{1-ab}=\tan c

Equations in inverse sines and cosines

Isolate, then take the sine

sin⁡−1u=θ ⇒ u=sin⁡θ,−π2≤θ≤π2\sin^{-1}u=\theta\ \Rightarrow\ u=\sin\theta,\quad -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}

When the domain pins x down

Squeezed argument

u≥1 and u≤1 ⇒ u=1u\ge1\ \text{and}\ u\le1\ \Rightarrow\ u=1

Common traps

Taking tangents loses the range

tan⁡(A+B)=1\tan(A+B)=1 also holds when A+B=−3π4A+B=-\frac{3\pi}4. Every root of the polynomial must be put back into the original equation; a negative root often fails.

Squaring adds roots

Taking a sine and squaring keeps every true root but can add false ones. In the worked example x=12x=\frac12 solves the quadratic but not the equation. Check each root in the original equation, with principal values.

Find the domain first

Manipulating the equation before checking the domain wastes time and can give roots at which a term is undefined. When a square root and an inverse sine share an argument, find the allowed xx first.

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