Question

PLEASE GIVE ME FAST ANSWER​

Answer 1

Step-by-step explanation:

From LHS .

$⇒ \frac{1}{secx - tanx \: } + \frac{1}{secx + tanx}$

$⇒ \frac{secx + tanx \: + secx - tanx}{sec {}^{2}x - tan {}^{2}x }$

$⇒ \frac{2secx}{sec {}^{2} x - tan {}^{2}x }$

We know that , sec²x-tan²x = 1

$⇒ \therefore \: 2secx$

But, secx = 1/cosx

• ∴ 2× 1/cosx

• = 2/cosx

• LHS = RHS proved .

Answer 2

Answer:

refer to this attachment ✅