Question

Prove that (1+secθ)/secθ=sin^2θ/(1-cosθ)

Answer 1

Use the identities,

$\sec ^2 \theta -1=\tan^2 \theta$

$(a+b)(a-b)=a^2-b^2$

Now,

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