Question

Question :-

If Cos^-1 C + Cos^-1 D = 3π/5 . Then Find the value of Sin^-1 C + Sin^-1 D . ✓

When I do this my answer is 2π/5 But it seems wrong to me .✓

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Answer 1

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

$\rm :\longmapsto\: {cos}^{ - 1}x + {cos}^{ - 1}y = \dfrac{3\pi}{5}$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\: {sin}^{ - 1}x + {sin}^{ - 1}y$

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

We know that,

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

Now,

Given that,

$\rm :\longmapsto\: {cos}^{ - 1}x + {cos}^{ - 1}y = \dfrac{3\pi}{5}$

$\rm :\longmapsto\:\dfrac{\pi}{2} - {sin}^{ - 1}x + \dfrac{\pi}{2} - {sin}^{ - 1}y = \dfrac{3\pi}{5}$

$\rm :\longmapsto\:\pi - {sin}^{ - 1}x - {sin}^{ - 1}y = \dfrac{3\pi}{5}$

$\rm :\longmapsto\:{sin}^{ - 1}x + {sin}^{ - 1}y = \pi - \dfrac{3\pi}{5}$

$\rm :\longmapsto\:{sin}^{ - 1}x + {sin}^{ - 1}y = \dfrac{5\pi - 3\pi}{5}$

$\bf :\longmapsto\:{sin}^{ - 1}x + {sin}^{ - 1}y = \dfrac{2\pi}{5}$

Additional Information :-

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

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

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

$\boxed{\rm :\longmapsto\: {tan}^{ -1}x + {tan}^{ - 1}y = {tan}^{ - 1}\bigg(\dfrac{x + y}{1 - xy} \bigg)}$

$\boxed{\rm :\longmapsto\: {tan}^{ -1}x - {tan}^{ - 1}y = {tan}^{ - 1}\bigg(\dfrac{x - y}{1 + xy} \bigg)}$

$\boxed{\rm :\longmapsto\: {sin}^{ -1}x - {sin}^{ - 1}y = {sin}^{ - 1}\bigg(x \sqrt{1 - {y}^{2}}- y \sqrt{1 - {x}^{2} } \bigg) }$

$\boxed{\rm :\longmapsto\: {sin}^{ -1}x + {sin}^{ - 1}y = {sin}^{ - 1}\bigg(x \sqrt{1 - {y}^{2}} + y \sqrt{1 - {x}^{2} } \bigg) }$

$\boxed{\rm :\longmapsto\: {cos}^{ -1}x +{cos}^{ - 1}y = {cos}^{ - 1}\bigg(xy - \sqrt{1 - {y}^{2}}\sqrt{1 - {x}^{2} } \bigg) }$

$\boxed{\rm :\longmapsto\: {cos}^{ -1}x - {cos}^{ - 1}y = {cos}^{ - 1}\bigg(xy + \sqrt{1 - {y}^{2}}\sqrt{1 - {x}^{2} } \bigg) }$