If 0- x - 1 - then tan-1-1-x21-x is equal to

The correct option is B 12cos^ 1x tan^ 1(√(1 x^2)1+x) Let x=cosθ =tan^ 1(√(1 cos^2θ)1+cosθ) tan^ 1sinθ1+cos^2θ2 sin^2θ2 =tan^ 12s ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

If 0< x < 1...

Question

If $0 < x < 1$ , then ${tan}^{- 1} (\frac{\sqrt{1 - x^{2}}}{1 + x})$ is equal to

\frac{1}{2} {cos}^{- 1} x

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{cos}^{- 1} \frac{\sqrt{1 + x}}{2}

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{sin}^{- 1} \sqrt{\frac{1 - x}{2}}

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\frac{1}{2} \sqrt{\frac{1 + x}{1 - x}}

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Solution

The correct option is B $\frac{1}{2} {cos}^{- 1} x$
${tan}^{- 1} (\frac{\sqrt{1 - x^{2}}}{1 + x})$
Let $x = cos θ$
$= {tan}^{- 1} (\frac{\sqrt{1 - {cos}^{2} θ}}{1 + cos θ})$
${tan}^{- 1} \frac{sin θ}{1 + {cos}^{2} \frac{θ}{2} - {sin}^{2} \frac{θ}{2}}$
$= {tan}^{- 1} \frac{2 sin \frac{θ}{2} cos \frac{θ}{2}}{2 {cos}^{2} \frac{θ}{2}}$
$= {tan}^{- 1} (tan \frac{θ}{2}) = \frac{θ}{2}$ = $\frac{1}{2} {cos}^{- 1} x$

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Similar questions

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
If

0 < x < 1

, then

\sqrt{1 + x^{2}} {[{x c o s ({c o t}^{- 1} x) + s i n ({c o t}^{- 1} x)}^{2} - 1]}^{1 / 2}

is equal to
(a)

\frac{x}{\sqrt{1 + x^{2}}}

(b)

x

(c)

x \sqrt{1 + x^{2}}

(d)

\sqrt{1 + x^{2}}

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

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
(a)

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

(b)

s i n [(1 / 5) {c o s}^{- 1} x] = 1

(c)

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

Solveis equal to

(A) (B) (C) (D)

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

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If 0<x<1 , then tan−1(√1−x21+x) is equal to

The correct option is B 12cos−1xtan−1(√1−x21+x)Let x=cosθ=tan−1(√1−cos2θ1+cosθ)tan−1sinθ1+cos2θ2−sin2θ2=tan−12sinθ2cosθ22cos2θ2=tan−1(tanθ2)=θ2=12cos−1x

If $0 < x < 1$ , then ${tan}^{- 1} (\frac{\sqrt{1 - x^{2}}}{1 + x})$ is equal to