Advertisements
Advertisements
प्रश्न
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
उत्तर
L.H.S. = `(1 + sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A)`
= `((1 + sin A)(cosec A + cot A) - (1 - sin A)(cosec A - cot A))/((cosec A - cot A)(cosec A + cot A))`
= `(cosec A + cot A + sin A cosec A + sin A cot A - cosec A + cot A + sin A cosec A - sin A cos A)/(cosec^2A - cot^2A)`
= 2 cot A + 2 sin A cosec A
= 2 cot A + 2 `1/(cosec A) xx cosec A`
= 2 (cot A + 1)
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove that:
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
`(1+ tan theta + cot theta )(sintheta - cos theta) = ((sec theta)/ (cosec^2 theta)-( cosec theta)/(sec^2 theta))`
Write True' or False' and justify your answer the following :
The value of the expression \[\sin {80}^° - \cos {80}^°\]
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
If 1 – cos2θ = `1/4`, then θ = ?
Prove that `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B
Prove that
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
