Question

solve this question given in the image​

Answer 1

Answer:

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Answer 2

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

$\rm :\longmapsto\: {sin}^{ - 1}\dfrac{x}{13} + {cosec}^{ - 1}\dfrac{13}{12} = \dfrac{\pi}{2}$

$\rm :\longmapsto\: {sin}^{ - 1}\dfrac{x}{13} = \dfrac{\pi}{2} - {cosec}^{ - 1}\dfrac{13}{12}$

We know,

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

Using this identity, we get

$\rm :\longmapsto\: {sin}^{ - 1}\dfrac{x}{13} ={sec}^{ - 1}\dfrac{13}{12}$

We know,

$\red{ \boxed{ \sf{ \: {sec}^{ - 1}x = {cos}^{ - 1}\dfrac{1}{x}}}}$

Using this identity, we get

$\rm :\longmapsto\: {sin}^{ - 1}\dfrac{x}{13} ={cos}^{ - 1}\dfrac{12}{13}$

We know,

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

Using this identity, we get

$\rm :\longmapsto\: {sin}^{ - 1}\dfrac{x}{13} ={sin}^{ - 1}\dfrac{ \sqrt{ {13}^{2} - {12}^{2} } }{13}$

$\rm :\longmapsto\: {sin}^{ - 1}\dfrac{x}{13} ={sin}^{ - 1}\dfrac{ \sqrt{169 - 144} }{13}$

$\rm :\longmapsto\: {sin}^{ - 1}\dfrac{x}{13} ={sin}^{ - 1}\dfrac{ \sqrt{25} }{13}$

$\rm :\longmapsto\: {sin}^{ - 1}\dfrac{x}{13} ={sin}^{ - 1}\dfrac{5}{13}$

On comparing, we get

$\bf\implies \:x = 5$

Additional Information :-

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

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

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

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

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

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