The value of sin-1 x - cos-1 x -x- - 1 is

The correct option is C π / 2 sin^ 1x+cos^ 1x sin^ 1x=θ_1 ⇒sinθ_1=x cos^ 1x=θ_2 ⇒cosθ_2=x tanθ_1=x√(1 x^2) tanθ_2=√(1 x^2)x ...

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Byju's Answer

Standard XII

Mathematics

Global Maxima

The value of ...

Question

The value of $s i n^{- 1} x + c o s^{- 1} x (| x | \geq 1)$ is

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π

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π / 2

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- π / 2

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Solution

The correct option is C $π / 2$
${sin}^{- 1} x + {cos}^{- 1} x$
${sin}^{- 1} x = θ_{1} \Rightarrow sin θ_{1} = x$
${cos}^{- 1} x = θ_{2} \Rightarrow cos θ_{2} = x$
$tan θ_{1} = \frac{x}{\sqrt{1 - x^{2}}}$
$tan θ_{2} = \frac{\sqrt{1 - x^{2}}}{x}$
$tan (θ_{1} + θ_{2}) = \frac{tan θ_{1} + tan θ_{2}}{1 - tan θ_{1} tan θ_{2}}$
$= \frac{\frac{x}{\sqrt{1 - x^{2}}} + \frac{\sqrt{1 - x^{2}}}{x}}{1 - 1}$
$= \frac{x^{2} + 1 - x^{2}}{x \sqrt{1 - x^{2}} (0)}$
$tan (θ_{1} + θ_{2}) = \infty$
$θ_{1} + θ_{2} = \frac{π}{2}$
$∴ {sin}^{- 1} x + {cos}^{- 1} x = \frac{π}{2}$

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Similar questions

Q. Match the column

	List I		List II
A.	Range of $f (x) = {sin}^{- 1} x + {cos}^{- 1} x + {cot}^{- 1} x$ is	1.	$[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$
B.	Range of $f (x) = cot - 1 x + {tan}^{- 1} x + c o s e c^{- 1} x$ is	2.	$[\frac{π}{2}, \frac{3 π}{2}]$
C.	Range of $f (x) = {cot}^{- 1} x + {tan}^{- 1} x + {cos}^{- 1} x$ is	3.	${0, π}$
D.	Range of $f (x) = {sec}^{- 1} x + c o s e c^{- 1} x + {sin}^{- 1} x$ is	4.	$(\frac{π}{2}, \frac{3 π}{2})$

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

Evaluate

(a)

{c o s}^{- 1} x + {c o s}^{- 1} [\frac{x}{2} + \frac{\sqrt{(3 - 3 x^{2})}}{2}]

(\frac{1}{2} \leq x \leq 1)

(b)

c o s (2 {c o s}^{- 1} x + {s i n}^{- 1} x)

x = 1 / 5

where

0 \leq {c o s}^{- 1} x \leq π

and

- π / 2 \leq {s i n}^{- 1} x \leq π / 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) Solve the equation :

{c o s}^{- 1} (\sqrt{6} x) + {c o s}^{- 1} (3 \sqrt{3} x^{2}) = \frac{π}{2}

(b) If

{s i n}^{1} 6 x + {s i n}^{- 1} 6 \sqrt{3} x = - π / 2

, then

x = . . . . . . .

(c)

{s i n}^{- 1} x + {s i n}^{- 1} 2 x = π / 3

Inverse circular functions,Principal values of ${sin}^{- 1} x, {c o s}^{- 1} x, {tan}^{- 1} x$ .
${tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}$ , $x y < 1$
$π + {tan}^{- 1} \frac{x + y}{1 - x y}$ , $x y > 1$ .
Evaluate the following :
(a) $sin [\frac{π}{3} - {sin}^{- 1} (- \frac{1}{2})]$
(b) $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$

Q. Assertion (

A

) lf

0 < x < \frac{π}{2}

then

{sin}^{- 1} (c o s x) + {cos}^{- 1} (s i n x) = π - 2 x

Reason (R)

{cos}^{- 1} x = \frac{π}{2} - {sin}^{- 1} x \forall x \in [0, 1]

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The value of sin−1x+cos−1x(|x|≥1) is

The correct option is C π/2sin−1x+cos−1xsin−1x=θ1⇒sinθ1=xcos−1x=θ2⇒cosθ2=xtanθ1=x√1−x2tanθ2=√1−x2xtan(θ1+θ2)=tanθ1+tanθ21−tanθ1tanθ2=x√1−x2+√1−x2x1−1=x2+1−x2x√1−x2(0)tan(θ1+θ2)=∞θ1+θ2=π2∴sin−1x+cos−1x=π2

The value of $s i n^{- 1} x + c o s^{- 1} x (| x | \geq 1)$ is