# 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- 2sin^{2}A)} / 2sinAcosA (Since, cos2x = 1 - 2sin^{2}x and sin2x = 2sinxcosx)

= (1-1 + 2sin^{2}A) / 2sinAcosA

= 2sin^{2}A /2sinAcosA

= sinA/cosA

= tanA

= RHS

**ii) 2sinAcos ^{3}A - 2sin^{3}AcosA = sin4A/2**

LHS = 2sinAcos^{3}A - 2sin^{3}AcosA

On taking 2sinAcosA common, we get

2sinAcosA × (cos^{2}A - sin^{2}A)

=sin2A × cos2A (Since, 2sinxcosx = sin2x and cos^{2}x - sin^{2}x = 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) 2sinAcos^{3}A - 2sin^{3}AcosA = sin4A/2

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

**Summary:**

Proof of i . cosec2A - cot2A = tanA ii. 2sinAcos^{3}A - 2sin^{3}AcosA = sin4A/2 is as shown above.

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