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प्रश्न
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove the following:
`secA/(secA + 1) + secA/(secA - 1) = 2"cosec"^2A`
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उत्तर
L.H.S. = `secA/(secA + 1) + secA/(secA - 1)`
= `(sec^2A - secA + sec^2A + secA)/(sec^2A - 1`
= `(2sec^2A)/tan^2A` ...(∵ sec2 A – 1 = tan2 A)
= `(2/cos^2A)/(sin^2A/cos^2A)`
= `2/sin^2A`
= 2 cosec2 A = R.H.S.
संबंधित प्रश्न
Prove the following identities:
sec2A + cosec2A = sec2A . cosec2A
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
