Prove the following i. cosec2A - cot2A = tanA ii. 2sinAcos3A - 2sin3AcosA = sin4A/2

Solution:

To Prove that cosec2A - cot2A = tanA, we will use trigonometric identities.

i) Cosec2A - Cot2A = TanA

LHS = cosec2A - cot2A

⇒ 1/sin2A - cos2A/sin2A   (Since, cosec x = 1/ sin x and cot x = cos x / sin x)

On taking LCM, we get

⇒ (1- cos2A)/sin2A

⇒ {1 - (1- 2sin²A)} / 2sinAcosA  (Since, cos2x = 1 - 2sin²x  and sin2x = 2sinxcosx)

= (1-1 + 2sin²A) / 2sinAcosA

= 2sin²A /2sinAcosA                            

= sinA/cosA                                                                                                                  

= tanA

= RHS

ii) 2sinAcos³A - 2sin³AcosA = sin4A/2

LHS = 2sinAcos³A - 2sin³AcosA

On taking 2sinAcosA common, we get

2sinAcosA × (cos²A - sin²A)

=sin2A × cos2A  (Since, 2sinxcosx = sin2x and cos²x - sin²x = cos2x)

On multiplying and dividing this with 2, we get

⇒(2/2) × sin2Acos2A

Substitute 2A with θ, we get

=(2/2) × sinθ cosθ                                                                                             

= (1/2) sin2θ  (Since, 2sinxcosx = sin2x)

Therefore, (1/2) sin2(2A) = (1/2) sin4A or sin4A/2

= RHS

Hence, Proved.

Thus,(i) cosec2A - cot2A = tanA. and (ii)  2sinAcos³A - 2sin³AcosA = sin4A/2

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Prove the following i. cosec2A - cot2A = tanA ii. 2sinAcos3A - 2sin3AcosA = sin4A/2

Summary:

Proof of i . cosec2A - cot2A = tanA  ii. 2sinAcos³A - 2sin³AcosA = sin4A/2 is as shown above.