Question

show that ( 1 + sinA/cosA × sinB/cosB) ² + ( sinA/cosA - sinB/cosB) ² = 1/cos²A.cos²B

pls answer soon as I am having preboard this week

I would prefer hand written answer than typed one

pls help by answering the above question​

Answer 1

Answer:

use sin/cos = tan then answer is very easy from there..

Answer 2

Step-by-step explanation:

${(1 + \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \times \frac{ \sin( \beta ) }{ \cos( \beta ) } )}^{2} + {( \frac{ \sin( \alpha ) }{ \cos( \alpha ){} } - \frac{ \sin( \beta ) }{ \cos( \beta ) } )}^{2} \\ \\ {(1 + \tan( \alpha ) \tan( \beta ) )}^{2} + {( \tan( \alpha ) - \tan( \beta )) }^{2} \\ \\ (1 + { \tan( \alpha ) }^{2} { \tan( \beta ) }^{2} + 2 \tan( \alpha ) \tan( \beta ) + { \tan( \alpha ) }^{2} + { \tan( \beta ) }^{2} - 2 \tan( \alpha ) \tan( \beta ) ) \\ \\ using \: idntity \\ 1 + { \tan( \alpha ) }^{2} = { \sec( \alpha ) }^{2} \\ \\ \\ \\ so \\ \\ 1 + (1 - { \sec( \alpha ) }^{2} )(1 - { \sec( \beta ) }^{2} ) + 2 - { \sec( \alpha ) }^{2} - { \sec( \beta ) }^{2} \\ 4 - 2( { \sec( \alpha ) }^{2} + { \sec( \beta ) }^{2} ) + { \sec( \alpha ) }^{2} { \sec( \beta ) }^{2}$

solve it further you will approch yo the answer

I think you are in class 10th if you are instersted then prepare for boards together