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Trigonometric Ratios of Compound Angles

Prove the fol...

Question

Prove the following identity : $(s i n A + s e c A)^{2}$ + $(c o s A + c o s e c A)^{2}$ = $(1 + s e c A . c o s e c A)^{2}$

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Solution

LHS
$= (s i n A + s e c A)^{2} + (c o s A + c o s e c A)^{2}$

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

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

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

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

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

$= (1 + s e c A c o s e c A)^{2}$ = RHS

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