Cos tan 1 x sin cot 1 3 4


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

Answered by admin @



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)}


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