Question

Prove that 1/sec0-tan0= sec0+tan0

Answer 1

Answer:

Step-by-step explanation:

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