Prove the following trigonometric identities:

$t a n^{2} A + c o t^{2} A = s e c^{2} A c o s e c^{2} A - 2$

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Solution

$L H S space equals space tan squared A space plus thin space co t squared A equals space fraction numerator sin squared A over denominator cos squared A end fraction plus fraction numerator cos squared A over denominator sin squared A end fraction equals fraction numerator sin to the power of 4 A plus cos to the power of 4 A over denominator sin squared A cos squared A end fraction equals fraction numerator open parentheses sin squared A plus cos squared A close parentheses squared minus 2 sin squared A cos squared A over denominator sin squared A cos squared A end fraction equals fraction numerator 1 minus 2 sin squared A cos squared A over denominator sin squared A cos squared A end fraction left parenthesis u sin g space sin squared A plus cos squared A space equals space 1 right parenthesis equals space s e c squared A cos e c squared A minus 2 space left parenthesis a s fraction numerator 1 over denominator sin A end fraction space equals space cos e c A space a n d space fraction numerator 1 over denominator cos A end fraction space equals space s e c A right parenthesis$
since LHS = RHS hence proved

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Similar questions

Prove the following trigonometric identities:

$\frac{t a n^{2} A}{1 + t a n^{2} A} + \frac{c o t^{2} A}{1 + c o t^{2} A} = 1$

\cos^{2} A + \frac{1}{1 + \cot^{2} A} = 1

\frac{\tan A}{{(1 + \tan^{2} A)}^{2}} + \frac{\cot A}{(1 + \cot^{2} A)} = \sin A \cos A

\frac{1 - \tan^{2} A}{\cot^{2} A - 1} = \tan^{2} A

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