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Question
Prove the following identities: cot θ - tan θ = `(2 cos^2 θ - 1)/(sin θ cos θ)`.
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Solution
LHS = cot θ - tan θ
= `cos θ/sin θ - sin θ/cos θ`
= `(cos^2 θ - sin^2 θ)/(sin θ. cos θ)`
= `(cos^2 θ - (1 - cos^2 θ))/(sin θ. cos θ)`
= `(2cos^2 θ - 1)/(sin θ. cos θ)`
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following identities:
`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
Prove the following identities:
`(1 + cos theta - sin^2 theta )/(sin theta (1 + cos theta)) = cot theta`
What is the value of 9cot2 θ − 9cosec2 θ?
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
Prove that cot2θ × sec2θ = cot2θ + 1.
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
