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Question
Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
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Solution
`cot^2A-cot^2B`
`=cos^2A/sin^2A-cos^2B/sin^2B`
`=(cos^2Asin^2B-cos^2Bsin^2A)/(sin^2Asin^2B)`
`=(cos^2A(1-cos^2B)-cos^2B(1-cos^2A))/(sin^2Asin^2B)`
`=(cos^2A-cos^2Acos^2B-cos^2B+cos^2Bcos^2A)/(sin^2Asin^2B)`
`=(cos^2A-cos^2B)/(sin^2Asin^2B)`
`=(1-sin^2A-1+sin^2B)/(sin^2Asin^2B)`
`=(-sin^2A+sin^2B)/(sin^2Asin^2B)`
`=sin^2B/(sin^2AsinB)-sin^2A/(sin^2Asin^2B)`
`=1/sin^2A-1/sin^2B`
= cosec2A - cosec2B
RELATED QUESTIONS
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following identities:
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
If tanθ `= 3/4` then find the value of secθ.
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
If x = a tan θ and y = b sec θ then
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
