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प्रश्न
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
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उत्तर
LHS = cosec2 (90° - θ) + cot2 (90° - θ)
= sec2 θ + tan2θ
= 1 + tan2θ + tan2θ
= 1 + 2 tan2θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Show that none of the following is an identity:
`sin^2 theta + sin theta =2`
Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
Prove that secθ + tanθ =`(costheta)/(1-sintheta)`.
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Prove that `(sin θ tan θ)/(1 - cos θ) = 1 + sec θ.`
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
Prove that `(cos^2θ)/(sinθ) + sin θ = "cosec" θ`.
