If x-sin -1 K- y-cos -1 K- -1- K- 1- then the correct relationship is-

The correct option is C x+y=π2 nbsp;x=sin ^ 1 K ,y=cos ^ 1 K nbsp;∵sin ^ 1 θ nbsp; +cos ^ 1 θ nbsp; =π nbsp;/ 2 nbsp;∴ x+y=sin ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Trigonometric Ratios of Standard Angles

If x=sin -1...

Question

If $x = {sin}^{- 1} K$ , $y = {cos}^{- 1} K$ , $- 1 \leq K \leq 1$ , then the correct relationship is-

x + y = 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x - y = 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x + y = \frac{π}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x - y = \frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $x + y = \frac{π}{2}$
$x = {sin}^{- 1} K, y = {cos}^{- 1} K$
$∵ {sin}^{- 1} θ + {cos}^{- 1} θ = \frac{π}{2}$
$∴ x + y = {sin}^{- 1} K + {cos}^{- 1} K = \frac{π}{2}$

Suggest Corrections

Similar questions

Q. If

{cos}^{- 1} x + {cos}^{- 1} y = \frac{π}{2}

, then prove that

{cos}^{- 1} x = {sin}^{- 1} y

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = π$ then $x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = k (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})$ where k is equal to

Q. If

s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = π

, then

x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = k (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})

, where

k

is equal to

Find the real part of $(\frac{(x + i y) + i}{(x + i y) + 2})$
Take $(| (x + 2) + i y |)$ = $\sqrt{\frac{1}{k}}$

Q. If

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{k} & x^{k + 2} & x^{k + 3} y^{k} & y^{k + 2} & y^{k + 3} z^{k} & z^{k + 2} & z^{k + 3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})

, then

Join BYJU'S Learning Program

If x=sin−1K, y=cos−1K, −1≤K≤1, then the correct relationship is-

The correct option is C x+y=π2x=sin−1K,y=cos−1K∵sin−1θ+cos−1θ=π2∴x+y=sin−1K+cos−1K=π2

If $x = {sin}^{- 1} K$ , $y = {cos}^{- 1} K$ , $- 1 \leq K \leq 1$ , then the correct relationship is-

The correct option is C $x + y = \frac{π}{2}$
$x = {sin}^{- 1} K, y = {cos}^{- 1} K$
$∵ {sin}^{- 1} θ + {cos}^{- 1} θ = \frac{π}{2}$
$∴ x + y = {sin}^{- 1} K + {cos}^{- 1} K = \frac{π}{2}$