Question

Derive: How?


`\cot(2a) = \frac{ {2cot}^(2)a - 1}{2cota}`
Please write all steps. [ using compound angle formula]
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Answer 1

see proof below

Explanation:

cot(2a) = (cot^2a - 1)/2cota

Remember that cotangent is the inverse of tan...

cot = 1/tan => cot(2a) = 1/tan2a

Here I used the formula for tan2a, or in other words tan2a = 2tana/1 - tan^2a

Therefore the inverse of tan2a, 1/tan2a, should be the inverse of 2tanA/1 - tan^2a, or 1 - tan^2a/2tana => 1/tan2a = 1 - tan^2a/2tana = cot(2a)

The reverse is true as well, tan is the inverse of cotangent...tanA = 1/cotA

cot(2a) = 1 - tan^2a/2tana = 1 - (1/cot^2a)/2(1/cota)

= cot^2a - 1/cot^2a * cota/2

= cot^2a - 1/2cota

L.H.S = R.H.S, hence proved

Answer 2

Answer: see proof below

Explanation:

Note: The given equation cannot be proven. Instead I will prove:

`\cot(2\alpha)=(\cot^2\alpha-1)/(2\cot \alpha)`

Use the Reciprocal Identity: cot A = (cos A)/(sin A)

Use the following Double Angle Identities:

cos (2A) = cos² A - sin² A

sin (2A) = 2 sin A · cos A

Proof LHS → RHS:

LHS: cot (2A)

`\text{Reciprocal:}\qquad \qquad (\cos (2\alpha))/(\sin (2\alpha))`

`\text{Double Angle:}\qquad \quad (\cos^2 \alpha-\sin^2 \alpha)/(2\sin \alpha \cdot \cos \alpha)`

`\text{Manipulate:}\qquad \quad (\cos^2 \alpha-\sin^2 \alpha)/(2\sin \alpha \cdot \cos \alpha)\bigg(((1)/(\sin^2 \alpha))/((1)/(\sin^2 \alpha))\bigg)`

`=((\cos^2 \alpha)/(\sin^2 \alpha)-(\sin^2 \alpha)/(\sin^2 \alpha))/((2\sin \alpa \cdot \cos \alpha)/(\sin \alpha \cdot \sin \alpha))`

`=(\cot^2 \alpha-1)/(2\cot \alpha)`

LHS = RHS
`\checkmark`