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Standard XII

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

The value of ...

Question

The value of ${sin}^{- 1} cos ({sin}^{- 1} x) + {cos}^{- 1} sin ({cos}^{- 1} x)$ is-

0

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\frac{π}{4}

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\frac{π}{2}

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π

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Solution

The correct option is D $\frac{π}{2}$
$∵ {sin}^{- 1} θ + {cos}^{- 1} θ = \frac{π}{2}$
${sin}^{- 1} (cos ({sin}^{- 1} x)) + {cos}^{- 1} (sin ({cos}^{- 1} x)) = {sin}^{- 1} (cos ({cos}^{- 1} \sqrt{1 - x^{2}}) + {cos}^{- 1} (sin ({sin}^{- 1} \sqrt{1 = x^{2}})))$
$= {sin}^{- 1} (\sqrt{1 - x^{2}}) + {cos}^{- 1} (\sqrt{1 - x^{2}}) = \frac{π}{2}$

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Similar questions

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

is equal to

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. 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 ({c o s}^{- 1} \frac{3}{5})

(b)

c o s ({t a n}^{- 1} \frac{3}{4})

and

c o s ({t a n}^{- 1} x)

s i n ({c o t}^{- 1} (1 + x)) = c o s ({t a n}^{- 1} x)

then x is
(a)

1 / 2

(b)

1

(c)

0

(d)

- 1 / 2

(d)

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

(e)

s i n (2 {s i n}^{- 1} 0.8)

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The value of sin−1cos(sin−1x)+cos−1sin(cos−1x) is-

The correct option is D π2∵sin−1θ+cos−1θ=π2sin−1(cos(sin−1x))+cos−1(sin(cos−1x))=sin−1(cos(cos−1√1−x2)+cos−1(sin(sin−1√1=x2)))=sin−1(√1−x2)+cos−1(√1−x2)=π2

The value of ${sin}^{- 1} cos ({sin}^{- 1} x) + {cos}^{- 1} sin ({cos}^{- 1} x)$ is-