Question

Prove:
cos theta/1+sin theta + cos theta/1-sin theta = 2sec theta
​

Answer 1

Step-by-step explanation:

$\frac{cos \theta}{1 +sin \theta} + \frac{cos \theta}{1 - sin \theta} = 2 \: sec \theta \: \\ LHS = \frac{cos \theta}{1 +sin \theta} + \frac{cos \theta}{1 - sin \theta} \: \\ = \frac{cos \theta(1 - sin \theta) + cos \theta(1 + sin \theta)}{(1 +sin \theta)(1 - sin \theta)} \\ = \frac{cos \theta -cos \theta sin \theta + cos \theta + cos \theta sin \theta }{1 - {sin}^{2} \theta } \: \\ = \frac{2 \: cos \theta}{ {cos}^{2} \theta} \\ = \frac{2 }{ {cos} \theta} \\ = 2 \: sec \theta \\ = RHS \\ thus \: proved.$

Answer 2

Answer and explanation:

To prove : $\frac{\cos\theta}{1+\sin\theta}+\frac{\cos\theta}{1-\sin\theta}=2\sec\theta$

Proof :

Taking LHS,

$LHS=\frac{\cos\theta}{1+\sin\theta}+\frac{\cos\theta}{1-\sin\theta}$

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

$LHS=\frac{\cos\theta-\cos\theta\sin\theta+\cos \theta+\cos \theta\sin \theta}{1^2-\sin^2\theta}$

$LHS=\frac{\cos\theta+\cos\theta}{1-\sin^2\theta}$

$LHS=\frac{2\cos\theta}{\cos^2\theta}$

$LHS=\frac{2}{\cos\theta}$

$LHS=2\sec\theta$

$LHS=RHS$

Hence proved.