Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
Advertisements
उत्तर
LHS = `(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ)`
= `((1 + sinθ)(cosecθ + cotθ) - (1 - sinθ)(cosecθ - cotθ))/(cosec^2θ - cot^2θ)`
= `(cosecθ + cotθ + 1 + cosθ - cosecθ + cotθ + 1 - cosθ)/(1 + cot^2θ - cot^2θ)` (∵ `cosec^2θ = 1 + cot^2θ`)
= 2 + 2cotθ = 2(1 + cotθ)
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`
Prove the following identities:
`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
Prove that:
`1/(sinA - cosA) - 1/(sinA + cosA) = (2cosA)/(2sin^2A - 1)`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
If x = asecθ + btanθ and y = atanθ + bsecθ , prove that `x^2 - y^2 = a^2 - b^2`
Prove that `(sin^2theta)/(cos theta) + cos theta` = sec θ
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
