Question

(1 - cos A )( 1 + sec A) = tan A sin A​

Answer 1

${ \boxed{\boxed{\begin{array}{c | c} \underline{ \bf \: process}& \underline{\bf \: explanation} \\ \\ \bf \: L.H.S=(1 - cosA)(1 + secA)& \bf \: given \\ \\ \bf = (1 - cosA)(1 + \frac{1}{cosA})& \bf \because \: secA = \frac{1}{cosA} \\ \\ \bf = (1 - cosA) \frac{(cosA + 1)}{cosA}& \\ \\ \bf = \frac{ {1}^{2} - {cos}^{2} A }{cosA} & \bf \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \\ \\ \bf = \frac{ {sin}^{2} A}{cosA} & \bf \because \: {sin}^{2} x + {cos}^{2}x = 1 \\ \\ \bf = \frac{sinA}{cosA} \times sinA& \: \\ \\ \bf = tanA.sinA& \bf \because \: tanx = \frac{sinx}{cosx} \\ \\ \bf = R.H.S\end{array}}}}$