Question

If

$sin^{-1}(x)+sin^{-1}(y)=\dfrac{2\pi}{3}$

Then what is the value of

$cos^{-1}(x)+cos^{-1}(y)$

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: {sin}^{ - 1}x + {sin}^{ - 1}y = \dfrac{2\pi}{3}$

We know,

$\rm :\longmapsto\: {sin}^{ - 1}x + {cos}^{ - 1}x = \dfrac{\pi}{2}$

So,

$\rm :\longmapsto\: {sin}^{ - 1}x = \dfrac{\pi}{2} - {cos}^{ - 1}x$

Now,

Given equation can be reduced as

$\rm :\longmapsto\:\dfrac{\pi}{2} - {cos}^{ - 1}x + \dfrac{\pi}{2} - {cos}^{ - 1}y = \dfrac{2\pi}{3}$

$\rm :\longmapsto\:\pi - {cos}^{ - 1}x - {cos}^{ - 1}y = \dfrac{2\pi}{3}$

$\rm :\longmapsto\:{cos}^{ - 1}x + {cos}^{ - 1}y =\pi - \dfrac{2\pi}{3}$

$\rm :\longmapsto\:{cos}^{ - 1}x + {cos}^{ - 1}y = \dfrac{3\pi - 2\pi}{3}$

$\rm :\longmapsto\:{cos}^{ - 1}x + {cos}^{ - 1}y = \dfrac{\pi }{3}$

More to know

$\rm :\longmapsto\: {sec}^{ - 1}x + {cosec}^{ - 1}x = \dfrac{\pi}{2}$

$\rm :\longmapsto\: {tan}^{ - 1}x + {cot}^{ - 1}x = \dfrac{\pi}{2}$

$\boxed{ \sf{ \: {sin}^{ - 1}x + {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} }}}$

$\boxed{ \sf{ \: {sin}^{ - 1}x - {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2} } - y \sqrt{1 - {x}^{2} }}}$