Question

Prove that:
$\frac{ \cosθ }{1- \sinθ } + \frac{ \cosθ }{1+ \sinθ } = 2 \secθ$
​

Answer 1

Step-by-step explanation:

$LHS =$

$\implies\frac{ \cosθ }{1- \sinθ } + \frac{ \cosθ }{1+ \sinθ } = 2 \secθ$

$\implies \frac{ \cos \theta(1 + \sin \theta) + \cos \theta(1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta) }$

$\frac{ \cos \theta(1 + \sin \theta) + \cos \theta(1 - \sin \theta)}{(1 - \sin ^{2} \theta) }$

$\implies \frac{ \cos \theta + \sin \theta \: cos \theta + cos \theta - \sin \theta \cos \theta}{ { \cos}^{2} \theta } \: \: \: \:$

$\:\:\:\:\: \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:[1 - { \sin}^{2} \theta = { \cos}^{2} \theta)]$

$\implies \frac{2 \not\cos \theta}{ { \cos}^{ \not2} \theta }$

$\implies2 \times \frac{1}{ \cos \theta}$

$\implies2 \sec \theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( \frac{1}{ \cos \theta} = \sec \theta)$

$RHS \: = 2 \sec \theta$

$Since \: \: LHS=RHS, \: \: hence \: \: proved$

Answer 2

$let \: theta \: be \: alpha \: ( \alpha ) \: \:$

$i \: am \: not \: finding \: theta$

$solution -$

$\frac{ \cos\alpha }{1 - \sin\alpha } + \frac{ \cos\alpha }{1 + \sin\alpha } = 2 \sec\alpha$

$taking \: LHS \:$

$\frac{ \cos\alpha (1 + \sin \alpha ) \: + \: \cos \alpha(1 - \sin \alpha )} {(1 - \sin \alpha ) (1 + \sin \alpha ) }$

$\frac{ \cos \alpha \: + \: (\sin\alpha \cos\alpha) \: + \: \cos \alpha \: - \: (\sin\alpha \cos \alpha )}{ {1}^{2} \: - \: \ \sin ^{2} \ \alpha }$

$\frac{ \cos \alpha \: + \: \cos \alpha }{ { \cos ^{2} \alpha } } \: = \: \frac{2 \cos\alpha }{ \cos^{2} \alpha } \: = \: \frac{2}{ \cos\alpha }$

$= 2 \sec\alpha$

$LHS \: = \: RHS \:$

$proved$

cheese

:)