Question

Prove that $\dfrac{1+\sec\theta}{\sec\theta}=\dfrac{\sin^2\theta}{1-\cos\theta}$.
 

Hint:

When proving trig identities, try converting everything to $\sin\theta$ and $\cos\theta$ and look for common factors like $(1-\cos\theta)$ or $(1+\cos\theta)$.

Answer 1

 

Step 1: Simplify the LHS
\[ \frac{1+\sec\theta}{\sec\theta}=\frac{1}{\sec\theta}+1=\cos\theta+1. \]

Step 2: Simplify the RHS using $\sin^2\theta=1-\cos^2\theta$
\[ \frac{\sin^2\theta}{1-\cos\theta}=\frac{1-\cos^2\theta}{1-\cos\theta} =\frac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta}=1+\cos\theta. \] Conclusion:
Both sides reduce to $1+\cos\theta$, hence proved.
 \[ \boxed{\dfrac{1+\sec\theta}{\sec\theta}=\dfrac{\sin^2\theta}{1-\cos\theta}} \]