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Visual SolutionHard
Question
If sin⁡−1(2a1+a2)+cos⁡−1(1−a21+a2)=tan⁡−1(2x1−x2)\sin^{-1}\left(\dfrac{2a}{1+a^2}\right) + \cos^{-1}\left(\dfrac{1-a^2}{1+a^2}\right) = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right), where a,x∈(0,1)a, x \in (0, 1), then the value of xx is:
(A)a1−a2\dfrac{a}{1-a^2}
(B)a2a^2
(C)2a1−a2\dfrac{2a}{1-a^2}
(D)a1+a2\dfrac{a}{1+a^2}
Solution Path
Recognize all three terms as double-angle forms of tan inverse. LHS = 4*tan inverse(a), RHS = 2*tan inverse(x). Solve: x = tan(2*tan inverse(a)) = 2a/(1-a^2).
01Question Setup
1/4
Given sin⁡−12a1+a2+cos⁡−11−a21+a2=tan⁡−12x1−x2\sin^{-1}\frac{2a}{1+a^2} + \cos^{-1}\frac{1-a^2}{1+a^2} = \tan^{-1}\frac{2x}{1-x^2} with a,x∈(0,1)a, x \in (0,1). Find xx.
x=  ?x = \;?
02Apply Double Angle IdentitiesKEY INSIGHT
2/4
All three terms are double-angle forms: sin⁡−12a1+a2=2tan⁡−1a\sin^{-1}\frac{2a}{1+a^2} = 2\tan^{-1}a, cos⁡−11−a21+a2=2tan⁡−1a\cos^{-1}\frac{1-a^2}{1+a^2} = 2\tan^{-1}a, and tan⁡−12x1−x2=2tan⁡−1x\tan^{-1}\frac{2x}{1-x^2} = 2\tan^{-1}x.
LHS=4tan⁡−1a,  RHS=2tan⁡−1x\text{LHS} = 4\tan^{-1}a,\; \text{RHS} = 2\tan^{-1}x
03Solve for x
3/4
From 4tan⁡−1a=2tan⁡−1x4\tan^{-1}a = 2\tan^{-1}x, we get tan⁡−1x=2tan⁡−1a\tan^{-1}x = 2\tan^{-1}a, so x=tan⁡(2tan⁡−1a)x = \tan(2\tan^{-1}a).
tan⁡−1x=2tan⁡−1a\tan^{-1}x = 2\tan^{-1}a
04Final Answer
4/4
Using tan⁡(2θ)=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta} with θ=tan⁡−1a\theta = \tan^{-1}a: x=2a1−a2x = \frac{2a}{1-a^2}.
2a1−a2\boxed{\dfrac{2a}{1-a^2}}
Concepts from this question2 concepts unlocked
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Double Angle Inverse Trig Identities

STANDARD

The identity 2*tan inverse(x) can be expressed as sin inverse(2x/(1+x^2)) when |x| <= 1, as cos inverse((1-x^2)/(1+x^2)) when x >= 0, and as tan inverse(2x/(1-x^2)) when |x| < 1. Similarly, sin inverse(2x*sqrt(1-x^2)) = 2*sin inverse(x) when |x| <= 1/sqrt(2), but equals pi - 2*sin inverse(x) when x > 1/sqrt(2). The domain restrictions are critical.

2tan⁡−1x=sin⁡−12x1+x2  (∣x∣≤1),sin⁡−1(2x1−x2)=2sin⁡−1x  (∣x∣≤12)2\tan^{-1}x = \sin^{-1}\frac{2x}{1+x^2} \;(|x| \le 1), \quad \sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x \;\left(|x| \le \frac{1}{\sqrt{2}}\right)

JEE loves testing whether students remember the domain restrictions. The clean formula works only in the specified range. Outside that range, you must use the alternate form. This is where most students lose marks.

Simplify expressions with substitutionDomain-restricted identitiesProve that expressions equal a constant
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Principal Value Range

EASY

Every inverse trigonometric function returns a unique value from a restricted range called the principal value range. For sin inverse: [-pi/2, pi/2], for cos inverse: [0, pi], for tan inverse: (-pi/2, pi/2). The principal value is the unique angle in this range whose trig ratio equals the given value. This restriction makes the inverse function single-valued.

sin⁡−1:[−1,1]→[−π2,π2],cos⁡−1:[−1,1]→[0,π],tan⁡−1:R→(−π2,π2)\sin^{-1}: [-1,1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \quad \cos^{-1}: [-1,1] \to [0, \pi], \quad \tan^{-1}: \mathbb{R} \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

The principal value range is the single most important concept in this chapter. Every JEE question on inverse trig ultimately tests whether you can bring the answer into the correct range. Getting this wrong means every subsequent step is also wrong.

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