Question

Please answer!!!!
Prove LHS=RHS

Answer 1

GIVEN :-

• √(1 + cos A)/(1 - cos A) = (1 + cos A)/sin A.

TO PROVE :-

• L.H.S = R.H.S.

SOLUTION :-

→ √(1 + cos A)/(1 - cos A) = (1 + cos A)/sin A

★ By Rationalizing the denominator in L.H.S Part,

→ √[(1 + cos A) (1 + cos A)/(1 - cos A) (1 + cos A)] = (1 + cos A)/sin A

→ √[( 1 + cos A)²/(1² - cos² A)] = (1 + cos A)/sin A.

→ [(1 + cos A)/√(1 - cos² A)] = (1 + cos A)/sin A.

★ By using identity:- 1 - cos² A = sin² A.

→ [(1 + cos A)/(√sin² A) = (1 + cos A)/sin A

→ (1 + cos A)/sin A = (1 + cos A)/sin A

L.H.S = R.H.S

❏ HENCE VERIFIED.

★ ADDITIONAL INFORMATION ★

→ sin² A + cos² A = 1.

→ tan² A = sin²A/cos²A.

→ cot² A = cos²A/sin²A.

Answer 2

Step-by-step explanation:

√(1 + cos A)/(1 - cos A) = (1 + cos A)/sin A

By Rationalizing the denominator in L.H.S Part,

⟹√[(1 + cos A) (1+ cos A)/(1 - cos A) (1 + cos A)] = (1 + cos A)/sin A

⟹√[(1+ cos A)²/(12 - cos ²A)] = (1 + cos A)/sin A.

⟹[(1+ cos A)/b(1 - cos ²A)] = (1 + cos A)/sin A.

By using identity:- 1- cos²A = sin²A.

⟹[(1+ cos A)/(√sin ²A) = (1 + cos A)/sin A

⟹(1 + cos A)/sin A = (1 + cos A)/sin A

LHS=RHS

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