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Trigonometric Identities

secA.cosec2A=...

Question

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

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Solution

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

Take the LHS:

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

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

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

Since, $t a n^{2} A + 1 = s e c^{2} A a n d c o t^{2} A + 1 = c o s e c^{2} A$

= $s e c^{2} A + c o s e c^{2} A$

= $\frac{1}{c o s^{2} A} + \frac{1}{S i n^{2} A}$

$\frac{(S i n^{2} A + C o s^{2} A)}{c o s^{2} A S i n^{2} A}$

= $\frac{1}{c o s^{2} A s i n^{2} A}$

$S e c^{2} A c o s e c^{2} A = R H S$

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Q. tan²A/1+tan²A + cot²A/1+cot²A=sec²Acos²A-2secAcosA

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

{sec}^{2} A {csc}^{2} A = {tan}^{2} A + {cot}^{2} A + 2

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

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

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