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प्रश्न
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
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उत्तर
RHS = `1 + 2 tan θ/cos θ + 2 tan^2 θ`
= `1 + 2 sin θ/cos^2θ + 2 sin^2 θ/cos^2 θ`
= `(cos^2 θ + 2sin θ + 2 sin^2 θ)/(cos^2θ)`
= `(1 - sin^2θ + 2 sin θ + 2 sin^2θ )/(1 - sin^2θ)`
= `(1 + sin^2θ + 2 sin θ)/(1 - sin^2θ)`
= `(1 + sin θ)^2/( 1 + sin θ)(1 - sin θ)`
= `(1 + sin θ)/(1 - sin θ)`
= LHS
Hence proved.
संबंधित प्रश्न
Express the ratios cos A, tan A and sec A in terms of sin A.
Prove the following trigonometric identities
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If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
`(sec^2 theta-1) cot ^2 theta=1`
`(sec^2 theta -1)(cosec^2 theta - 1)=1`
The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is
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`sinθ(1 + tanθ) + cosθ(1 +cotθ) = secθ + cosecθ`
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If x = h + a cos θ, y = k + b sin θ.
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.
