Question

Prove that:
(See the attachment below)

Answer 1

Hope this helps u...

Answer 2

Answer:

Step-by-step explanation:

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

$L.H.S\\\\=\frac{1}{sec\theta-tan\theta}-\frac{1}{cos\theta}\\\\=\frac{sec^2\theta-tan^2\theta}{sec\theta-tan\theta}-sec\theta\\\\=\frac{(sec\theta-tan\theta)(sec\theta+tan\theta)}{sec\theta-tan\theta}-sec\theta\\\\=sec\theta+tan\theta-sec\theta\\\\=tan\theta............(1)$

$R.H.S\\\\=\frac{1}{cos\theta}-\frac{1}{sec\theta+tan\theta}\\\\=sec\theta-\frac{sec^2\theta-tan^2\theta}{sec\theta+tan\theta}\\\\=sec\theta-\frac{(sec\theta-tan\theta)(sec\theta+tan\theta)}{sec\theta+tan\theta}\\\\=sec\theta-(sec\theta-tan\theta)\\\\=sec\theta-sec\theta+tan\theta\\\\=tan\theta.............(2)$

From (1) and (2)

L..H.S=R.H.S