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प्रश्न
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
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उत्तर
R.H.S. = `(1 - sinA)/(1 + sinA)`
= `(1 - 1/(cosecA))/(1 + 1/(cosecA))`
= `(cosecA - 1)/(cosecA + 1)`
= `(cosecA - 1)/(cosecA + 1) xx (cosecA + 1)/(cosecA + 1)`
= `(cosec^2A - 1)/(cosecA + 1)^2 = cot^2A/(cosecA + 1)^2` ...(∵ cosec2 A – 1 = cot2 A)
= L.H.S.
संबंधित प्रश्न
If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p2 – 1) = 2p
Prove the following trigonometric identities.
`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
`(1-cos^2theta) sec^2 theta = tan^2 theta`
If `( sin theta + cos theta ) = sqrt(2) , " prove that " cot theta = ( sqrt(2)+1)`.
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
Let α, β be such that π < α – β < 3π. If sin α + sin β = `-21/65` and cos α + cos β = `-27/65`, then the value of `cos (α - β)/2` is ______.
