If x-1 x2-1 - 0- then find the value of sin 1-2tan-12x-1-x2-tan-1x

The correct option is C 1Given, #160; ( x 1 ) ( x^2+1 ) gt; 0 #160; #160; ⇒ #160; #160; x gt; 1 ∴ #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If x-1 x2+...

Question

If $(x - 1) (x^{2} + 1) > 0$ , then find the value of $sin (\frac{1}{2} {tan}^{- 1} \frac{2 x}{1 - x^{2}} - {tan}^{- 1} x)$

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Solution

The correct option is C -1
Given, $(x - 1) (x^{2} + 1) > 0$ $\Rightarrow$ $x > 1$
$∴$ $sin [\frac{1}{2} {tan}^{- 1} \frac{2 x}{1 - x^{2}} - {tan}^{- 1} x] = sin [\frac{1}{2} (- π + 2 {tan}^{- 1} x) - {tan}^{- 1} x]$

$= sin (- \frac{π}{2})$

$= - 1$

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Similar questions

Q. Prove the following

(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1

(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0

(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1

Solve the following equations for x:
(i) tan⁻¹2x + tan⁻¹3x = nπ + $\frac{3 π}{4}$

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹ $\frac{8}{31}$

(iii) $\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}$

(iv) sin⁻¹x + sin⁻¹2x = $\frac{π}{3}$

(v) $3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}$

(vi) $\cos^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6}$

(vii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x

(viii) tan (cos⁻¹x) = sin $(\cot^{- 1} \frac{1}{2})$

(ix) tan⁻¹ $(\frac{1 - x}{1 + x}) - \frac{1}{2}$ tan⁻¹x = 0, where x > 0

(x) cot⁻¹x − cot⁻¹(x + 2) = $\frac{π}{12}$ , x > 0

(xi) $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0$

(xii) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹ $(\frac{8}{79})$ , x > 0

(xiii) $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$