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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
Prove the following identities:
`cosecA + cotA = 1/(cosecA - cotA)`
Prove that:
`tanA/(1 - cotA) + cotA/(1 - tanA) = secA "cosec" A + 1`
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
Prove that `(tan^2"A")/(tan^2 "A"-1) + (cosec^2"A")/(sec^2"A"-cosec^2"A") = (1)/(1-2 co^2 "A")`
If x = r sin θ cos Φ, y = r sin θ sin Φ and z = r cos θ, prove that x2 + y2 + z2 = r2.
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)` = sec θ + tan θ
