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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

sin -1 cos si...

Question

${sin}^{- 1} (cos ({sin}^{- 1} x)) + {cos}^{- 1} (sin ({cos}^{- 1} x))$ is equal to

\frac{π}{4}

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\frac{π}{2}

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\frac{3 π}{4}

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0

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Solution

The correct option is B $\frac{π}{2}$
We have ${sin}^{- 1} (cos ({sin}^{- 1} x)) + {cos}^{- 1} (sin ({cos}^{- 1} x))$
$= {sin}^{- 1} [cos (\frac{π}{2} - {cos}^{- 1} x)] + {cos}^{- 1} [sin (\frac{π}{2} - {sin}^{- 1} x)]$
$= {sin}^{- 1} [sin ({cos}^{- 1} x)] + {cos}^{- 1} [cos ({sin}^{- 1} x)]$
$= {sin}^{- 1} x + {cos}^{- 1} x = \frac{π}{2}$

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Similar questions

Q. If

α = {sin}^{- 1} (cos ({sin}^{- 1} x))

and

β = {cos}^{- 1} (sin ({cos}^{- 1} x))

, then

tan α \cdot tan β

is equal to

Q. For

0 \leq {cos}^{- 1} x \leq π

and

- \frac{π}{2} \leq {sin}^{- 1} x \leq \frac{π}{2}

, the value of

cos ({sin}^{- 1} x + 2 {cos}^{- 1} x)

x = \frac{1}{5}

is:

Q. The value of

{sin}^{- 1} cos ({sin}^{- 1} x) + {cos}^{- 1} sin ({cos}^{- 1} x)

is-

Q. Evaluate the following:
(i)

\cos (\sin^{- 1} \frac{3}{5})

(ii)

\sin (\cos^{- 1} \frac{4}{5})

(iii)

\cos (\sin^{- 1} - \frac{3}{5})

(iv)

\tan (\cos^{- 1} \frac{8}{17})

(v)

cosec \{\cos^{- 1} (- \frac{12}{13})\}

(vi)

\tan \{2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}\}

(vii)

\tan \frac{1}{2} (\cos^{- 1} \frac{\sqrt{5}}{3})

(viii)

\sin (\frac{1}{2} \cos^{- 1} \frac{4}{5})

(ix)

\cos (\sin^{- 1} \frac{3}{5} + \sin^{- 1} \frac{5}{13})

(x) sin (tan⁻¹ x + cot⁻¹ x)

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)

{s i n}^{- 1} (3 x / 5) + {s i n}^{- 1} (4 x / 5) = {s i n}^{- 1} x

(b)

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

a \leq {t a n}^{- 1} (\frac{1 - x}{1 + x}) \leq b

where

0 \leq x \leq 1

then

(a, b) =

(a)

(0, π)

(b)

(0, π / 4)

(c)

(- π / 4, π / 4)

(d)

(π / 4, π / 2)

(d) If

a \leq {({s i n}^{- 1} x)}^{3} + {({c o s}^{- 1} x)}^{3} \leq b

then (a,b) is equal to

(\frac{π^{3}}{32}, \frac{7 π^{3}}{8})

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sin−1(cos(sin−1x))+cos−1(sin(cos−1x)) is equal to

The correct option is B π2We have sin−1(cos(sin−1x))+cos−1(sin(cos−1x))=sin−1[cos(π2−cos−1x)]+cos−1[sin(π2−sin−1x)]=sin−1[sin(cos−1x)]+cos−1[cos(sin−1x)]=sin−1x+cos−1x=π2

${sin}^{- 1} (cos ({sin}^{- 1} x)) + {cos}^{- 1} (sin ({cos}^{- 1} x))$ is equal to