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प्रश्न
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
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उत्तर
LHS = `(sec^2θ - sin^2θ)/tan^2θ`
= `(1/cos^2θ - sin^2θ)/(sin^2θ/cos^2θ)`
= `(1 - sin^2θcos^2θ)/((cos^2θ)/(sin^2θ/cos^2θ))`
= `(1 - sin^2θcos^2θ)/sin^2θ`
= `1/sin^2θ - (sin^2θcos^2θ)/(sin^2θ)`
= `cosec^2θ - cos^2θ`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`cosec theta sqrt(1 - cos^2 theta) = 1`
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
`sin^2 theta + 1/((1+tan^2 theta))=1`
If 5 `tan theta = 4,"write the value of" ((cos theta - sintheta))/(( cos theta + sin theta))`
If `sqrt(3) sin theta = cos theta and theta ` is an acute angle, find the value of θ .
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
Prove the following identities.
`(sin^3"A" + cos^3"A")/(sin"A" + cos"A") + (sin^3"A" - cos^3"A")/(sin"A" - cos"A")` = 2
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
