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प्रश्न
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
पर्याय
tan2 θ
sec2 θ
1
–1
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उत्तर
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is –1.
Explanation:
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`
= `(cos^2 θ - 1)/(sin^2 θ)`
= `(-sin^2 θ)/(sin^2 θ)` ...(∵ sin2θ = 1 – cos2θ)
= –1
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
if `cosec theta - sin theta = a^3`, `sec theta - cos theta = b^3` prove that `a^2 b^2 (a^2 + b^2) = 1`
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
Prove that sin( 90° - θ ) sin θ cot θ = cos2θ.
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
If tan α + cot α = 2, then tan20α + cot20α = ______.
