Question

Answer this question!!

Answer 1

$\frac{ { \cosec}^{2}a + { \sec }^{2}a }{ {cosec}^{2} a - \ {sec}^{2}a } = \frac{1 + {tan}^{2}a }{1 - { tan }^{2} a} \\ lhs.. \\ \frac{ { \cosec}^{2}a + { \sec }^{2}a }{ {cosec}^{2} a - \ {sec}^{2}a } \\ = \frac{ \frac{1}{ {sin}^{2}a } + \frac{1}{ {cos}^{2} a}}{ \frac{1}{ {sin}^{2}a } - \frac{1}{ {cos}^{2} a} } \\ = \frac{ { \cos }^{2} a + {sin}^{2} a}{ {sin}^{2}a \times {cos}^{2} a} \div \frac{ {cos}^{2} a - {sin}^{2} a}{ {sin}^{2} a \times {cos}^{2} a} \\ = \frac{1}{ {cos}^{2}a - {sin}^{2}a } \: \: \: ({{sin}^{2} a \times {cos}^{2} a} \: is \: cancelled \: and \: { {cos}^{2} a + {sin}^{2} a} \: = 1)$
$rhs.. \\ \frac{1 + { \tan}^{2}a }{1 - { \tan }^{2}a} \\ = \frac{1 + \frac{ {sin}^{2}a }{ {cos}^{2} a} }{1 - \frac{ {sin}^{2}a }{ {cos}^{2} a}} \\ = \frac{ {cos}^{2}a + {sin}^{2} a }{ {cos}^{2} a} \div \frac{ {cos}^{2}a - {sin}^{2} a }{ {cos}^{2} a} \\ = \frac{1}{ {cos}^{2} a} \times \frac{ {cos}^{2} a}{ {cos}^{2} a - {sin}^{2}a } \: \: \: \: ( {cos}^{2} a \: is \: cancelled \: ) \\ = \frac{1}{ {cos}^{2}a - {sin}^{2} a}$
Hence ,, LHS = RHS
-------------------------------------------------------------
HOPE THIS WILL HELP YOU DEAR.