Question

Prove that sin A tan A \ 1 - cos A = 1 + cos A

Answer 1

Answer:

Step-by-step explanation:(sinA*sinA/cosA) /(1-cosA)

sin^2A/(cosA (1-cosA ))

(1-cos^2A)/ (cosA (1-cosA))

(1+cosA)(1-cosA) / (cosA (1-cosA))

(1+cosA)/cosA

(1/cosA)+(cosA/cosA)

secA+1

1+secA proved

Answer 2

Answer:

$\frac{ \sin( \alpha ) \tan( \alpha ) }{1 - \cos( \alpha ) } \\ = > \frac{ \sin( \alpha ) \frac{ \sin( \alpha ) ) }{ \ \cos ( \alpha ) } }{1 - \cos( \alpha ) } \\ = > \frac{ { \sin ^{2} ( \alpha ) } }{ \cos( \alpha ) (1 - \cos( \alpha ) )} \\ = > \frac{1 - { \cos ^{2} ( \alpha ) } }{ \cos( \alpha )(1 - \cos( \alpha )) } \\ = > \frac{(1 + \cos( \alpha ))(1 - \cos( \alpha ) ) }{ \cos( \alpha ) (1 - \cos( \alpha )) } \\ \\ here \: (1 - \cos( \alpha ) ) \: is \: cancel \: \\ = > (1 + \cos( \alpha ) ) \times \sec( \alpha )$

IN RHS WE HAVE TO PROVE

1+COS@ • SEC@ OR sec@+1

Formula used :

1/cos@=sec@

tan@=sin@/cos@

HOPE THIS HELPS YOU☺️☺️