Question

Prove that Cos A/1- sin A + Cos A/ 1+ sin A= 2sec A...good answer will mark as brainlist​

Answer 1

Step-by-step explanation:

Given :-

Cos A/(1- sin A) + Cos A/( 1+ sin A)

To Find :-

Cos A/(1- sin A) + Cos A/ (1+ sin A) = 2sec A.

Solution :-

On taking LHS in the equation

Cos A/(1- sin A) + Cos A/ (1+ sin A)

=> Cos A [ 1/(1-Sin A) + 1/(1+Sin A)]

=>CosA[{(1+SinA)+(1-SinA)}/(1-SinA)(1+SinA)]

=> Cos A [ (1+Sin A +1-Sin A)/(1²-Sin²A)]

Since (a+b)(a-b) = a²-b²

Where, a = 1 and b = Sin A

=> Cos A [ (1+1)/(1-Sin² A)]

=> Cos A [ 2/(1-Sin² A)]

We know that

Sin² A + Cos² A = 1

=> Cos A (2/Cos² A)

=> 2Cos A / Cos² A

=> 2 Cos A /(Cos A × Cos A)

On cancelling Cos A in both the numerator and the denominator then

=> 2/Cos A

=> 2(1/Cos A)

=> 2 Sec A

=> RHS

=> LHS = RHS

Hence, Proved.

Answer :-

Cos A/(1- sin A) + Cos A/(1+ sin A) =2secA.

Used Identities :-

→ (a+b)(a-b) = a²-b²

→ Sin² A + Cos² A = 1

→ 1/Cos A = Sec A

Answer 2

Step-by-step explanation:

$\frac{cosA}{1-sinA} +\frac{cosA}{1+sinA} = cosA[ \frac{1}{1-sinA} +\frac{1}{1+sinA}]\\\\=cosA\frac{(1+sinA)+(1-sinA)}{(1+sinA)(1-sinA)} \\\\=\frac{2cosA}{1-sin^2A} \\\\=\frac{2cosA}{cos^2A}\\\\\=\frac{2}{cosA}\\\\=2secA$