Question

Sin A + Cot A / Sin A - Cos A + Sin A - Cos A / Sin A + Cot A =. 2 / Sin ^2 A - Cos ^2 A. = 2 / 2 Sin ^2 A - 1

Answer 1

Correct Question

→ (sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(sin²A - cos²A) = 2/(2sin²A - 1)

We Know That,

• sin²A + cos²A = 1
• cos²A = 1 - sin²A
• (a + b)(a - b) = a² - b²
• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• (a + b)² + (a - b)² = 2(a² + b²)

Proof

(Given in attachment)

Procedure

1. Take L.C.M of L.H.S and then use above identities.

2. Break down cos²A into sin²A by using distinct identity.

3. You will get final values as required.

Hence Proved

Answer 2

★Answer:

★Proved!★

★Step-by-step explanation:

★Correction in your question:

(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(sin²A - cos²A) = 2/(2sin²A - 1)

★Solution:

Consider L.H.S:                         (Left hand side)

$\frac{Sin A+cosA}{SinA-cosA} +\frac{SinA-cosA}{SinA+cosA} =\frac{(SinA+cosA)^2+(SinA-cosA)^2}{(SinA-CosA)(SinA+cosA)}$

$=\frac{Sin^2A+cos^2A+2sinA\ cosA+sin^2A+cos^2A-2sinA\ cosA}{(sin^2A-cos^2A)}$

$=\frac{1+1}{(sin^2A-cos^2A)}$

$=\frac{2}{(Sin^2A-cos^2A)}$

$=\frac{2}{[sin^2A-(1-sin^2A)]}$

$=\frac{2}{[Sin^2A-1+sin^2A]}$

$=\frac{2}{2sin^2A-1}$

=R.H.S                 (Right hand side)

★Some Important Identies:★

★sin²A + cos²A = 1

★cos²A = 1 - sin²A

★(a + b)(a - b) = a² - b²

★(a + b)² = a² + b² + 2ab

★(a - b)² = a² + b² - 2ab

★(a + b)² + (a - b)² = 2(a² + b²)

★We also use these identies to find L.H.S , R.H.S of expressions.★

★Note:★

★Don't forget to use BRACKETS!!★

★Try to post full&correct question.★

★Hope my answer helps you.★