Question

(tan a + sec a )² - (tan a - sec a)² = 4 sin a / 1 - sin² a​

Answer 1

$\bf\green{\underline{ }}lhs = ( {tan \alpha + sec \alpha )}^{2} - ( {tan \alpha - sec \alpha )}^{2} \\ \\ = > \bf\green{\underline{ {a}^{2} - {b}^{2} = (a + b)(a - b) }} \\ ......\bf\red{ a = tan \alpha + sec \alpha } \\ ......\bf\red{ b = tan \alpha - sec \alpha } \\ \\ \bf\green{\underline{ }} = > (tan \alpha + sec \alpha + tan \alpha - sec \alpha ) \\ \bf\green{\underline{ }} \: \: \: \: \: \: \: \: (tan \alpha + sec \alpha - tan \alpha + sec \alpha ) \\ \\ \bf\green{\underline{ }} = > (2tan \alpha )(2sec \alpha ) \\ \\ \bf\large\pink{\underline{ = > 4tan \alpha .sec \alpha }} - - - (1) \\ \\ \\ \\ \bf\large\green{\underline{ }}rhs = \frac{4sin \alpha }{1 - {sin}^{2} \alpha } \\ \\ = > \bf\large\green{\underline{ {sin}^{2} \alpha + {cos}^{2} \alpha = 1 }} \\ \\ \bf\large\green{\underline{ }} = > \frac{4sin \alpha }{ {cos}^{2} \alpha } \\ \\ = > \bf\large\green{\underline{ }} \frac{4sin \alpha }{cos \alpha \times cos \alpha } \\ \\ \bf\large\green{\underline{ }} = > \frac{4sin \alpha }{cos \alpha } \times \frac{1}{cos \alpha } \\ \\ \bf\large\green{\underline{ \frac{sin \alpha }{cos \alpha } = tan \alpha }} \\ \\ = > \bf\large\green{\underline{ }}4tan \alpha \times \frac{1}{cos \alpha } \\ \\ \bf\large\green{\underline{ \frac{1}{cos \alpha } = sec \alpha }} \\ \\ = > \bf\large\pink{\underline{ 4tan \alpha .sec \alpha }} - - - (2)$

By equation (1) and (2) ,

LHS = RHS (verified)