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प्रश्न
Prove the following identities:
`cosecA + cotA = 1/(cosecA - cotA)`
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उत्तर
L.H.S. = `cosecA + cotA`
= `(cosecA + cotA)/1 xx (cosecA - cotA)/(cosecA - cotA)`
= `(cosec^2A - cot^2A)/(cosecA - cotA)`
= `(1 + cot^2A - cot^2A)/(cosecA - cotA)`
= `1/(cosecA - cotA)` = R.H.S.
संबंधित प्रश्न
(secA + tanA) (1 − sinA) = ______.
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
Prove the following identities:
`("cosec" theta + cot theta)/("cosec" theta - cot theta) = ("cosec" theta + cot theta )^2 = 1 + 2 cot^2 theta + 2 "cosec" theta cot theta`
Write the value of cosec2 (90° − θ) − tan2 θ.
Prove the following identities:
`(sec"A"-1)/(sec"A"+1)=(sin"A"/(1+cos"A"))^2`
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
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 5 tan β = 4, then `(5 sin β - 2 cos β)/(5 sin β + 2 cos β)` = ______.
