Advertisements
Advertisements
प्रश्न
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Advertisements
उत्तर
L.H.S. `=(cosec A)/(cosecA - 1) + (cosecA)/(cosecA + 1)`
= `(cosec A (cosec A + 1) + cosec A (cosec A - 1))/((cosec A - 1) (cosec A + 1))`
= `(cosec^2 A+cosec A + cosec^2 A-cosec A)/((cosec A)^2 - (1)^2)`
= `(2 cosec^2 A)/(cosec^2 A - 1)`
= `(2 cosec^2 A)/(cot^2 A)` ...(∵ cosec2 A – 1 = cot2 A)
= `2(1/cancel(sin^2A))/(cos^2A/cancel(sin^2A))`
= `2/cos^2A`
= 2 sec2 A
= R.H.S.
संबंधित प्रश्न
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
If `cos B = 3/5 and (A + B) =- 90° ,`find the value of sin A.
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1.
Prove that `"cosec" θ xx sqrt(1 - cos^2theta)` = 1
Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B
If cosA + cos2A = 1, then sin2A + sin4A = 1.
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
