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प्रश्न
Prove the following identity :
`tan^2A - sin^2A = tan^2A.sin^2A`
उत्तर
LHS = `tan^2A - sin^2A`
= `sin^2A/cos^2A - sin^2A = (sin^2A(1 - cos^2A))/cos^2A`
= `sin^2A/cos^2A.sin^2A = tan^2A.sin^2A` = RHS
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संबंधित प्रश्न
If secθ + tanθ = p, show that `(p^{2}-1)/(p^{2}+1)=\sin \theta`
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Choose the correct alternative:
sin θ = `1/2`, then θ = ?
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
Prove that sec2θ – cos2θ = tan2θ + sin2θ
Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B
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.
