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Question
Prove the following identities:
`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`
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Solution
L.H.S. = tan A – cot A
= `(sin A)/(cos A) - (cos A)/(sin A)`
= `(sin^2A - cos^2A)/(sin A cos A)`
= `(1 - cos^2A - cos^2A)/(sin A cos A)` ...(∵ sin2A = 1 – cos2A)
= `(1 - 2cos^2A)/(sin A cos A)`
= R.H.S.
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tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
