Question

$\Huge\bf\maltese{\underline{\green{Answer°᭄}}}\maltese$
$\frac{ \cos \: \alpha }{1 + sin \: \alpha } + \frac{1 + sin \: \alpha }{ \cos \: \alpha } = 2sec \alpha$

Please Answer Correctly. ​

Answer 1

Given Question of Trignometry:

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

Solution:

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

$\frac{ \cos \: \alpha }{1 + sin \: \alpha } + \frac{1 + sin \: \alpha }{ \cos \: \alpha } = 2sec \alpha \\ \frac{ {cos}^{2} \alpha + 1 + {sin}^{2} \alpha + 2 \: sin \: \alpha }{cos \: \alpha \: (1 + sin \: \alpha )} \\ \frac{2 + 2 \: sin \alpha \: }{cos \: \alpha \: (1 + sin \: \alpha ) } \\ \frac{2 \: (1 + \: sin \: \alpha )}{cos \: \alpha (1 + \: sin \: \alpha )} \\ \frac{2}{cos \: \alpha } \\ \\ 2 \: sec \: \alpha \\ \\ as \: we \: know \: that \: \frac{1}{cos \: \alpha } = sec \: \alpha \\ and \: therefore \: \frac{2}{cos \: \alpha } = 2sec \: \alpha \\ \\ LHS = RHS \\ Hence Proved$

Answer 2

$\dag\:\underline{\boxed{\bf{\orange{\dfrac{cos\alpha}{1+sin\alpha}+\dfrac{1+sin\alpha}{cos\alpha}}=\gray{2sec\alpha}}}}$

LHS :

$\sf:\implies\:\dfrac{cos\alpha}{1+sin\alpha}+\dfrac{1+sin\alpha}{cos\alpha}$

By cross multiplying,

$\sf:\implies\:\dfrac{[cos\alpha(cos\alpha)]+[(1+sin\alpha)(1+sin\alpha)]}{(1+sin\alpha)cos\alpha}$

$\sf:\implies\:\dfrac{cos^2\alpha+(1+sin\alpha)^2}{(1+sin\alpha)cos\alpha}$

We know that, (A + B)² = A² + 2AB + B²

$\sf:\implies\:\dfrac{cos^2\alpha+1+2\sin\alpha+sin^2\alpha}{(1+sin\alpha)cos\alpha}$

• sin²α + cos²α = 1

$\sf:\implies\:\dfrac{2+2sin\alpha}{(1+sin\alpha)cos\alpha}$

$\sf:\implies\:\dfrac{2(1+sin\alpha)}{(1+sin\alpha)cos\alpha}$

$\sf:\implies\:\dfrac{2}{cos\alpha}$

We know that, 1 / cos α = sec α

$\bf:\implies\:2sec\alpha=RHS$

Hence proved!!