Question

(sina - coseca)^2 + (cosa - seca) ^2 = tan^2a + cot^a -1 prove that​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

$1+ { \cot}^{2} a = { \cosec}^{2} a$

2.

$1 + { \tan}^{2} a = { \sec}^{2} a$

TO PROVE

${( \sin a - \cosec a)}^{2} + {( \cos a - \sec a)}^{2} = { \cot}^{2} a + { \tan}^{2} a - 1$

PROOF

${( \sin a - \cosec a)}^{2} + {( \cos a - \sec a)}^{2}$

$= { \sin}^{2} a - 2 \times \sin a \times \cosec a + { \cosec}^{2} a + { \cos}^{2} a - 2 \times \cos a \times \sec a + { \sec}^{2} a$

$= { \sin}^{2} a - 2 + { \cosec}^{2} a + { \cos}^{2} a - 2 + { \sec}^{2} a$

$= { \sin}^{2} a + { \cosec}^{2} a + { \cos}^{2} a + { \sec}^{2} a - 4$

$= { \sin}^{2} a + { \cos}^{2} a + { \cot}^{2} a + 1 + { \tan}^{2} a + 1- 4$

$= 1+ { \cot}^{2} a + 1 + { \tan}^{2} a + 1- 4$

$= { \cot}^{2} a + { \tan}^{2} a - 1$