Question

Sin inverse X + Sin inverse Y barabar to 2by 3 then Cos inverse X + Cos inverse Y equal to​

Answer 1

we know, $sin^{-1}A+cos^{-1}A=\frac{\pi}{2}$

so, $sin^{-1}X+cos^{-1}X=\frac{\pi}{2}$

$\implies sin^{-1}X=\frac{\pi}{2}-cos^{-1}X$........(1)

and $sin^{-1}Y+cos^{-1}Y=\frac{\pi}{2}$

$\implies sin^{-1}Y=\frac{\pi}{2}-cos^{-1}Y$........(2)

given, $sin^{-1}X+sin^{-1}Y=\frac{2}{3}$

from equations (1) and (2),

$\frac{\pi}{2}-cos^{-1}X+\frac{\pi}{2}-cos^{-1}Y=\frac{2}{3}$

or, $-cos^{-1}X-cos^{-1}Y+\pi=\frac{2}{3}$

or, $cos^{-1}X+cos^{-1}Y=\pi-\frac{2}{3}$

hence, $cos^{-1}X+cos^{-1}Y=\pi-\frac{2}{3}$

Answer 2

Answer:

$cos^{-1}x+cos^{-1}y=\frac{\pi}{3}$

Step-by-step explanation:

Formula used:

1.$sin^{-1}x$ is an angle whose sine value is x

That is

$If \:sin\theta=x, then \:sin^{-1}x=\theta$

$2.sin^{-1}x+cos^{-1}x=\frac{\pi}{2}$

$Given:\\\\sin^{-1}x+sin^{-1}y=\frac{2\pi}{3}\\\\(\frac{\pi}{2}-cos^{-1}x)+(\frac{\pi}{2}-cos^{-1}y)=\frac{2\pi}{3}\\\\\pi-(cos^{-1}x+cos^{-1}y)=\frac{2\pi}{3}\\\\cos^{-1}x+cos^{-1}y=\pi-\frac{2\pi}{3}\\\\cos^{-1}x+cos^{-1}y=\frac{3\pi-2\pi}{3}\\\\cos^{-1}x+cos^{-1}y=\frac{\pi}{3}$