Cos-1x-tan-12x-1-x2- then-

Puttting θ= cos^ 1x in RHS, we have R.H.S =tan^ 1√(1 x^2)/x =tan^ 1√(1 cos^2θ)/cosθ(0≤θ≤π) =tan^ 1sinθ/cosθ=tan^ 1tanθ =θ =cos^ 1x when π/2 lt;θ lt;π/2 i,e ...

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Standard XI

Mathematics

Properties Derived from Trigonometric Identities

cos-1x=tan-12...

Question

$c o s^{- 1} x = t a n^{- 1} \frac{2 x}{1 + x^{2}}$ , then:

x ϵ R

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- 1 x \leq 1, x \neq 0

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0 < x \leq 1

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- 1 \leq x < 0

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Solution

The correct option is C $0 < x \leq 1$

Puttting $θ = c o s^{- 1} x$ in RHS, we have
R.H.S $= t a n^{- 1} \frac{\sqrt{1 - x^{2}}}{x}$
$= t a n^{- 1} \frac{\sqrt{1 - c o s^{2} θ}}{c o s θ} (0 \leq θ \leq π)$
$= t a n^{- 1} \frac{s i n θ}{c o s θ} = t a n^{- 1} t a n θ$
$= θ = c o s^{- 1} x$ when $- \frac{π}{2} < θ < \frac{π}{2}$
i,e., $w h e n 0 \leq θ \leq \frac{π}{2}$
i.e., when $0 < c o s θ \leq 1$ i.e., $0 < x \leq 1$

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Similar questions

Q. If

{sin}^{- 1} \frac{2 x}{1 + x^{2}}; {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}; {tan}^{- 1} \frac{2 x}{1 - x^{2}} .

each is equal to

2 {tan}^{- 1} x .

,then show that

\int 2 {tan}^{- 1} x = 2 [x {tan}^{- 1} x - \frac{1}{2} log (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}

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

, then:

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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cos−1x=tan−12x1+x2, then:

The correct option is C 0<x≤1 Puttting θ=cos−1x in RHS, we have R.H.S =tan−1√1−x2x =tan−1√1−cos2θcosθ(0≤θ≤π) =tan−1sinθcosθ=tan−1tanθ =θ=cos−1x when −π2<θ<π2 i,e., when0≤θ≤π2 i.e., when 0<cosθ≤1 i.e., 0<x≤1

$c o s^{- 1} x = t a n^{- 1} \frac{2 x}{1 + x^{2}}$ , then: