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

Putting 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=tan^ 1 tan θ =θ =cos^ 1x when π/2 ...

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Standard XI

Mathematics

Basic Inverse Trigonometric Functions

cos-1x=tan-1√...

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

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. 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}

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

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

, then:

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) If

{t a n}^{- 1} \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt (1 + x^{2}) + \sqrt (1 + x^{2})} = α

. then prove that

x^{2} = s i n 2 α

(b) If

\frac{m t a n (α - θ)}{{c o s}^{2} θ} = \frac{n t a n θ}{{c o s}^{2} (α - θ)}

, then prove that

θ = \frac{1}{2} [α - {t a n}^{- 1} (\frac{n - m}{n + m} t a n α)]

(c)

{c o s}^{- 1} \frac{c o s α + c o s β}{1 + c o s α c o s β} = 2 {t a n}^{- 1} (t a n \frac{α}{2} t a n \frac{β}{2})

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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: