Prove that $\frac{1}{s e c A + t a n A} = s e c A - t a n A$

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Solution

Given LHS = $\frac{1}{s e c A - t a n A}$

On rationalizing we get,

= $\frac{1}{s e c A - t a n A} \times \frac{s e c A + t a n A}{s e c A + t a n A}$

= $\frac{s e c A + t a n A}{(s e c A - t a n A) (s e c A + t a n A)}$

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

= $\frac{s e c A + t a n A}{1}$

= $s e c A + t a n A$
LHS = RHS.

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