Cos -1 x -tan -1-1- x 2- x- then-A- None of theseB- X - R C- - x - 1- x-0D- 0-x - 1

The correct option is D 0 lt; x ≤ 1Putting θ=cos^ 1x in R.H.S., we have R.H.S =tan^ 1√(1 x^2)/x =tan^ 1√(1 cos^2 θ)/cos θ (0 ≤θ≤π) =tan^ 1sin θ/cos θ ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

cos -1 x =tan...

Question

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

X \in R

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- \leq x \leq 1, x \neq 0

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0 < x \leq 1

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N o n e o f t h e s e

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Solution

The correct option is C $0 < x \leq 1$
Putting $θ = c o s^{- 1} x$ in R.H.S., 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., when $0 \leq θ < \frac{π}{2}$
i.e., when $0 < c o s θ \leq 1$ i.e., $0 < x \leq 1$ .

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Similar questions

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

, then:

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

, then:

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

, then:

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

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cos−1x=tan−1√1−x2x, then:

The correct option is C 0<x≤1Putting θ=cos−1x in R.H.S., 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., when 0≤θ<π2 i.e., when 0<cosθ≤1 i.e., 0<x≤1.

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