Question

(sin A - cos A)/(sin A + cos A)+(sin A + cosA)/(sin A - cos A) =2/2 sin² A - 1​

Answer 1

Step-by-step explanation:

Refer to attachment.

The steps which you might see are lost in details, are the hints given below :-

Look at these identities :-

• sin²a + cos²a = 1
• cos²a = 1 - sin²a

So, when (-) was applied it changed the sign of denominator.

Other identities:-

• tan²a + 1 = sec²a
• cot²a +1 = cosec²a
• sina = 1/coseca
• cosa = 1/seca
• tana = sina/cosa $.$

Answer 2

Given :

(sin A - cos A)/(sin A + cos A)+

(sin A +cosA)/(sin A - cos A) =2/2 sin² A-1

Explain

Sin A+cosA/sin A-cos A+sin A-cos A/sin A+cosA= 2/sin²A-cos²A-2/1-2cos ²A

LHS Sin A+cosA/sin A-cos A+sin A-cos A/ sin A+cosA

((Sin A+cosA)² + (Sin A - cosA)²)/(sin A -cosA)(sinA+cosA)

={(sin²A+cos²A + 2sinA.cosA) + (sin²A+cos²A - 2sinA.cosA)]/(sin² A - cos²A)

= {(1+2sinA.cosA) + (1-2sinA.cosA) / (sin² A - cos²A)

[sin²0+cos²0=1]

= 1 + 1/ (sin² A - cos²A)

LHS = 2 / (sin² A - cos²A) = 2/(1- cos²A - cos²A) = 2/(1-2cos²A) = RHS

[sin²0 = 1 - cos²0]

$\: \:$