Question

Solve the following equation for x:
cos(tan-1 x )=sin(cot-1 3/4)​

Answer 1

Basic Identities Used :-

$\boxed{ \sf \:cos( {cos}^{ - 1}x) = x \: if \: x \in \: [- 1,1]}$

$\boxed{ \sf \:sin({sin}^{ - 1}x) = x \: if \: x \in \: [- 1,1]}$

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

Given that

$\rm :\longmapsto\:cos( {tan}^{ - 1}x) = sin\bigg( {cot}^{ - 1}\dfrac{3}{4} \bigg)$

$\rm :\longmapsto\:cos\bigg( {cos}^{ - 1}\dfrac{1}{ \sqrt{1 + {x}^{2} } } \bigg) = sin\bigg( {sin}^{ - 1}\dfrac{4}{5} \bigg)$

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{1 + {x}^{2} } } = \dfrac{4}{5}$

On squaring both sides, we get

$\rm :\longmapsto\:\dfrac{1}{1 + {x}^{2} } = \dfrac{16}{25}$

$\rm :\longmapsto\:25 = 16 + 16 {x}^{2}$

$\rm :\longmapsto\:25 - 16 = 16 {x}^{2}$

$\rm :\longmapsto\:9 = 16 {x}^{2}$

$\rm :\longmapsto\: {x}^{2} = \dfrac{9}{16}$

$\bf\implies \:x = \: \pm \: \dfrac{3}{4}$

Additional Information :-

$\boxed{ \sf \: {sin}^{ - 1}x + {cos}^{ - 1}x = \dfrac{\pi}{2} \: \: if \: x \in \: [ - 1,1]}$

$\boxed{ \sf \: {tan}^{ - 1}x + {cot}^{ - 1}x = \dfrac{\pi}{2} \: \: if \: x \in \: ( - \infty , \infty )}$

$\boxed{ \sf \: {sec}^{ - 1}x + {cosec}^{ - 1}x = \dfrac{\pi}{2} \: \: if \: x \in \: ( - \infty , \infty )}$

$\boxed{ \sf \: {tan}^{ -1 }x + {tan}^{ - 1}y = {tan}^{ - 1}\bigg(\dfrac{x + y}{1 - xy}\bigg) \: \: if \: xy \leqslant 1}$

$\boxed{ \sf \: {tan}^{ -1 }x + {tan}^{ - 1}y =\pi + {tan}^{ - 1}\bigg(\dfrac{x + y}{1 - xy}\bigg) \:if \: xy > 1}$

$\boxed{ \sf \:2 {tan}^{ - 1}x = {tan}^{ - 1}\bigg(\dfrac{2x}{1-{x}^{2}}\bigg)}$

$\boxed{ \sf \:2 {tan}^{ - 1}x = {sin}^{ - 1}\bigg(\dfrac{2x}{1 + {x}^{2}}\bigg)}$

$\boxed{ \sf \:2 {tan}^{ - 1}x = {cos}^{ - 1}\bigg(\dfrac{1 - {x}^{2} }{1 + {x}^{2}}\bigg)}$