Question

SecQ + tanQ) (1-sinQ)
=
cos​

Answer 1

Answer:

(secQ + tanQ)(1-sinQ) = cosQ

LHS = (1/cosQ +sinQ/cosQ)(1-sinQ)

=( 1+sinQ/cosQ)(1-sinQ)

= 1-sin^2Q/cosQ

=cos^2Q/cosQ

= cosQ

hope this will help you

Answer 2

$\huge \red{\mathfrak{solution}}$

To prove :

$\boxed{ \bold{(sec \theta + tan \theta)( 1 - sin \theta) = cos \theta}}$

LHS

$\leadsto \: ( \sec \theta +tan \theta)(1 - sin \theta) \\ \\ \leadsto \: ( \frac{1}{cos \theta} + \frac{sin \theta}{cos \theta} )(1 - sin \theta) \\ \\ \leadsto \: ( \frac{1 + sin \theta}{cos \theta} )(1 - sin \theta) \\ \\ \leadsto \: \frac{1 - {sin}^{2} \theta }{cos \theta} \\ \\ \leadsto \: \frac{ {cos}^{2} \theta }{cos \theta} \\ \\ \leadsto \cancel \frac{ {cos}^{2} \theta}{cos \theta} cos \theta \\ \\ \leadsto \: cos \theta$

LHS= RHS

$\huge \boxed { \green{ \bold{proved}}}$

FORMULA USED:

$\tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \\ \\ 1 - {sin}^{2} \alpha = { \cos}^{2} \alpha$