Math, asked by kanangupta738, 4 months ago

prove that cos²(tan^-1 2) +sin²(cot^-1 3) = 3/10​

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Answered by shadowsabers03
4

Let,

\displaystyle\sf{\longrightarrow \alpha=\tan^{-1}2}

\displaystyle\sf{\longrightarrow \tan\alpha=2}

\displaystyle\sf{\longrightarrow \sqrt{\sec^2\alpha-1}=2}

\displaystyle\sf{\longrightarrow \sec^2\alpha=5}

\displaystyle\sf{\longrightarrow \cos^2\alpha=\dfrac {1}{5}}

Let,

\displaystyle\sf{\longrightarrow \beta=\cot^{-1}3}

\displaystyle\sf{\longrightarrow \cot\beta=3}

\displaystyle\sf{\longrightarrow\sqrt{\csc^2\beta-1}=3}

\displaystyle\sf{\longrightarrow \csc^2\beta=10}

\displaystyle\sf{\longrightarrow \sin^2\beta=\dfrac {1}{10}}

Then,

\displaystyle\sf{\longrightarrow \cos^2(\tan^{-1}2)+\sin^2(\cot^{-1}3)=\cos^2\alpha+\sin^2\beta}

\displaystyle\sf{\longrightarrow \cos^2(\tan^{-1}2)+\sin^2(\cot^{-1}3)=\dfrac {1}{5}+\dfrac {1}{10}}

\displaystyle\sf{\longrightarrow\underline {\underline {\cos^2(\tan^{-1}2)+\sin^2(\cot^{-1}3)=\dfrac {3}{10}}}}

Hence Proved!

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