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Trigonometric Identity- 1

Question 5 ix...

Question

Question 5 (ix)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(i x) (c o s e c A - s i n A) (s e c A - c o s A) = \frac{1}{(t a n A + c o t A)}$
[Hint : Simplify L.H.S and R.H.S separately]

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Solution

$(i x) (c o s e c A - s i n A) (s e c A - c o s A) = \frac{1}{(t a n A + c o t A)}$
$L . H . S = (c o s e c A - s i n A) (s e c A - c o s A) = (\frac{1}{s i n A} - s i n A) (\frac{1}{c o s A} - c o s A) = [\frac{(1 - s i n^{2} A)}{s i n A}] [\frac{(1 - c o s^{2} A)}{c o s A}] = (\frac{c o s^{2} A}{s i n A}) \times (\frac{s i n^{2} A}{c o s A}) = c o s A s i n A R . H . S = \frac{1}{(t a n A + c o t A)} = \frac{1}{(\frac{s i n A}{c o s A} + \frac{c o s A}{s i n A})} = \frac{1}{[\frac{(s i n^{2} A + c o s^{2} A)}{s i n A c o s A}]} = c o s A s i n A L . H . S = R . H . S$

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