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Cot54degree/tan 36degree+tan20degree/cot70gegree-2

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Cot54degree/tan 36degree+tan20degree/cot70gegree-2

Posted by Deepa Prakash 5 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Rajan Kumar Pasi 5 years, 6 months ago

Q. {tex}\huge {\frac{\cot\ 54^\circ}{\tan\ 36^\circ}+\frac{\tan\ 20^\circ}{\cot\ 70^\circ} -2=?}\\{/tex}
Sol:

{tex}\huge {=> \ \frac{\cot\ 54^\circ}{\tan\ 36^\circ}+\frac{\tan\ 20^\circ}{\cot\ 70^\circ} -2}\\.\\ \huge {=> \ \frac{\tan\ 36^\circ}{\tan\ 36^\circ}+\frac{\tan\ 20^\circ}{\tan\ 20^\circ} -2}\\.\\ \huge {=> \ 1+1 -2}\\.\\ \huge {=> 0}\\.\\{/tex}

{tex}\Large {\Bigg( {\because\ \cot \ \theta=\tan\ (90^\circ-\theta)\\ \therefore \cot\ 54^\circ=\tan\ (90^\circ-54^\circ)=\tan\ 36^\circ\\ and,\ \cot\ 70^\circ=\tan\ (90^\circ-70^\circ)=\tan\ 20^\circ\Bigg)\\}}{/tex}

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