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प्रश्न
Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.
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उत्तर
LHS = sec2 θ + cosec2 θ
= `1/(cos^2 θ) + 1/(sin^2 θ)`
= `(sin^2 θ + cos^2 θ)/(sin^2 θ. cos^2 θ)`
= `1/(sin^2 θ. cos^2 θ)`
= `1/(sin^2 θ) xx 1/(cos^2 θ)`
= sec2 θ cosec2 θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
Prove the following trigonometric identities.
if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`
Prove that:
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
`1+(tan^2 theta)/((1+ sec theta))= sec theta`
`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`
Prove the following identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
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
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
If tan θ – sin2θ = cos2θ, then show that sin2 θ = `1/2`.
