Question

please prove that.. Urgent need of the answer.​

Answer 1

Proof:

$\text{LHS} = {\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta - 1}}\\ \\ \implies {\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta - 1}}\times {\frac{\sin\theta+\cos\theta+1}{\sin\theta+\cos\theta + 1}}\\ \\ \implies {\frac{(\sin\theta+1)^2-(\cos\theta)^2}{(\sin\theta+\cos\theta)^2 - 1^2}}\\  \\ \implies \large{\frac{\sin^2\theta+2\sin\theta-\cos^2\theta+1}{\sin^2\theta+\cos^2\theta+2\sin\thetacos\theta - 1}}\\ \\ \text{We know that:}\ \sin^2\theta + \cos^2\theta=1$

$\implies \frac{\sin^2\theta+2\sin\theta-\cos^2\theta +\sin^2\theta+\cos^2\theta}{2\sin\theta\cos\theta} \\ \\ \implies \frac{2\sin^2\theta+2\sin\theta}{2\sin\theta\cos\theta} \\ \\ \implies \frac{1+\sin\theta}{\cos\theta}\\ \\ \implies \frac{1}{\sec\theta}+\tan\theta = RHS$

Hence Proved.

Hope this helps.