Question

pls solve it properly urgent no unnecessary answers​

Answer 1

Given Question :-

Solve for x :-

$\rm :\longmapsto\:sin\bigg[ {sin}^{ - 1} \dfrac{1}{5} + {cos}^{ - 1}x\bigg] = 1$

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

Given Inverse Trigonometric function is

$\rm :\longmapsto\:sin\bigg[ {sin}^{ - 1} \dfrac{1}{5} + {cos}^{ - 1}x\bigg] = 1$

can be rewritten as

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

$\rm :\longmapsto\:{sin}^{ - 1} \dfrac{1}{5} + {cos}^{ - 1}x= {sin}^{ - 1}\bigg[sin\dfrac{\pi}{2} \bigg]$

We know,

$\boxed{ \tt{ \: {sin}^{ - 1}(sinx) = x \: }}$

So, using this, we get

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

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

We know,

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

So, using this, we get

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

$\bf\implies \:x \: = \: \dfrac{1}{5}$

More to know :-

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

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

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

$\boxed{ \tt{ \: {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1}\bigg[\dfrac{x + y}{1 - xy} \bigg] \: }}$

$\boxed{ \tt{ \: {tan}^{ - 1}x - {tan}^{ - 1}y = {tan}^{ - 1}\bigg[\dfrac{x - y}{1 + xy} \bigg] \: }}$

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

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