\(\frac{tan \ A}{1 + sec \ A} - \frac{tan \ A}{1 - sec \ A} = 2 \ cosec \ A\)
Proof:
LHS \(= \frac{tan \ A}{1 + sec \ A} - \frac{tan \ A}{1 - sec \ A}\)
\(=\)
\(=\) [Using \(a^2 - b^2 = (a + b)(a - b)\)]
\(=\) [Using \(1 + tan^2 \ \theta = sec^2 \ \theta\)]
\(=\)
\(=\)
\(=\)
\(= 2 \ cosec \ A\)
\(=\) RHS
Thus, \(\frac{tan \ A}{1 + sec \ A} - \frac{tan \ A}{1 - sec \ A} = 2 \ cosec \ A\).
Hence, we proved.
Answer variants:
\(\frac{2sec \ A}{tan \ A}\)
\(\frac{2}{cos \ A} \times \frac{cos \ A}{sin \ A}\)
\(\frac{tan \ A - tan \ A \ sec \ A - tan \ A - tan \ A \ sec \ A}{1 - sec^2 \ A}\)
\(\frac{2}{sin \ A}\)
\(\frac{-2 tan \ A \ sec \ A}{-tan^2 \ A}\)
\(\frac{tan \ A(1 - sec \ A) - tan \ A(1 + sec \ A)}{(1 + sec \ A)(1 - sec \ A)}\)