Question

prove that tan^2 theta+cot^2 theta=sec^2 theta cosec^2 theta-2

Answer 1

Answer:

$\ {tan}^{2} a + {cot}^{2} a = \frac{ {sin}^{2} a}{ {cos}^{2} a} + \frac{ {cos}^{2}a }{ {sin}^{2}a }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ {( {sin}^{2} a)}^{2} + { ({cos}^{2}a) }^{2} }{ { {cos}^{2}a } \: \: \ {sin}^{2}a }$

$since (\: {\: {a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab})$

$\: \: \: \: \: \: \: \: \: \: \: \: = \frac{ ({sin}^{2}a + {cos}^{2}a) - 2 {sin}^{2} a \: \: {cos}^{2} a}{ {cos}^{2} a \: \: {sin}^{2} a }$

$since (\: {sin}^{2} a + {cos}^{2} a = 1)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1 - 2 {sin}^{2} a \: \: {cos}^{2} a}{ {cos}^{2}a \: \: {sin}^{2} a}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{ {cos}^{2}a \: \: {sin}^{2}a } - \frac{2 {sin}^{2}a \: \: {cos}^{2}a }{ {cos}^{2} a \: \: {sin}^{2}a }$

$since( { \frac{1}{ {cos}^{2} a} } = {sec}^{2}a \: \: \\ and \frac{1}{ {sin}^{2}a } = {cosec}^{2} a)$

${tan}^{2}a+{cot}^{2}a={sec}^{2}a \:{cosec}^{2}a-2$