If sin^-12a1+a^2-cos^-11-b^21+b^2=tan^-12x1-x^2, then x is equal to

D. a-b/1+ab. Step 1: For suitable a,b,x: sin^-12a1+a^2=2tan^-1a, cos^-11-b^21+b^2=2tan^-1b and tan^-12x1-x^2=2tan^-1x. Step 2: The equation becomes 2tan^-1a-2tan^-1b=2tan^-1x, so tan^-1x=tan^-1a-tan^-1b. Step 3: x=tan(tan^-1a-tan^-1b)=a-b/1+ab. Check: Take b=a: the left side is 0, so x=0; option (D) gives 0, while (C) gives 2a1-a^2≠ 0. Trap: (C) a+b/1-ab uses the addition formula, treating the minus sign between the inverse functions as a plus.

Question 11 of 21 · Inverse Trigonometric Functions

Questionmedium

This question was previously asked in

NIMCET 2010

If $s in^{- 1} \frac{2 a}{1 + a ^{2}} - cos^{- 1} \frac{1 - b ^{2}}{1 + b ^{2}} = t an^{- 1} \frac{2 x}{1 - x ^{2}}$ , then $x$ is equal to

Answer (Detailed Solution Below)View answer ▾

Answer: Option D — $\frac{a - b}{1 + ab}$

Detailed Solution

For suitable $a, b, x$ : $s in^{- 1} \frac{2 a}{1 + a ^{2}} = 2 t an^{- 1} a$ , $cos^{- 1} \frac{1 - b ^{2}}{1 + b ^{2}} = 2 t an^{- 1} b$ and $t an^{- 1} \frac{2 x}{1 - x ^{2}} = 2 t an^{- 1} x$ .

The equation becomes $2 t an^{- 1} a - 2 t an^{- 1} b = 2 t an^{- 1} x$ , so $t an^{- 1} x = t an^{- 1} a - t an^{- 1} b$ .

$x = t an (t an^{- 1} a - t an^{- 1} b) = \frac{a - b}{1 + ab}$ .

✓ Check

Take

b = a

: the left side is

0

, so

x = 0

; option (D) gives

0

, while (C) gives

\frac{2 a}{1 - a ^{2}} \neq = 0

⚠ Common trap

(C)

\frac{a + b}{1 - ab}

uses the addition formula, treating the minus sign between the inverse functions as a plus.

Quick tip

$tan^{- 1} x = tan^{- 1} a - tan^{- 1} b \Rightarrow x = \frac{a - b}{1 + ab}$

💬 Discussions on this question

Ask a doubt ›

Still stuck on this question? Ask our faculty and fellow students — answers appear here.

Ask a doubt

Inverse Trigonometric Functions — Preparation

This question tests your understanding of Inverse Trigonometric Functions, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

Inverse trigonometric functions, such as sin^-1 x, cos^-1 x and tan^-1 x, return the angle whose trigonometric ratio is a given value. Because trigonometric functions repeat, each inverse is defined only on a principal value branch. For example, sin^-1 x takes values between -pi/2 and pi/2. This topic covers domains and principal value ranges, finding principal values, and key properties such as sin^-1 x + cos^-1 x = pi/2 and the formulas for tan^-1 x + tan^-1 y. Exam questions test simplifying expressions, solving equations, and evaluating compositions such as sin(cos^-1 x). The most common mistake is ignoring the principal range, for example assuming sin^-1(sin x) always equals x. Practise the inverse trigonometric functions questions below, each with a step-by-step solution.

‹ All Inverse Trigonometric Functions questions Browse all topics ›

If $s in^{- 1} \frac{2 a}{1 + a ^{2}} - cos^{- 1} \frac{1 - b ^{2}}{1 + b ^{2}} = t an^{- 1} \frac{2 x}{1 - x ^{2}}$ , then $x$ is equal to

💬 Discussions on this question

More Inverse Trigonometric Functions Questions

More Quantitative Aptitude Questions

Inverse Trigonometric Functions — Preparation

If sin−11+a22a​−cos−11+b21−b2​=tan−11−x22x​, then x is equal to

💬 Discussions on this question

More Inverse Trigonometric Functions Questions

More Quantitative Aptitude Questions

Inverse Trigonometric Functions — Preparation

If $s in^{- 1} \frac{2 a}{1 + a ^{2}} - cos^{- 1} \frac{1 - b ^{2}}{1 + b ^{2}} = t an^{- 1} \frac{2 x}{1 - x ^{2}}$ , then $x$ is equal to