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प्रश्न
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
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उत्तर
LHS = cosec2(90° - θ) - tan2 θ
LHS = sec2 θ - tan2 θ = 1
RHS = cos2(90° - θ) + cos2 θ
RHS = sin2θ + cos2 θ = 1
Hence, LHS = RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
`sqrt((1 - sinA)/(1 + sinA)) = cosA/(1 + sinA)`
Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
\[\frac{1 + \tan^2 A}{1 + \cot^2 A}\]is equal to
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
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.
