Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
उत्तर
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
LHS = `(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1)`
= `(cosec^2A + cosecA + cosec^2A - cosecA)/(cosec^2A - 1)`
= `(2cosec^2A)/cot^2A(Q cosec^2A - 1 = cot^2A)`
= `(2/sin^2A)/(cos^2A/sin^2A) = 2/cos^2A = 2sec^2A`
APPEARS IN
संबंधित प्रश्न
(secA + tanA) (1 − sinA) = ______.
if `a cos^3 theta + 3a cos theta sin^2 theta = m, a sin^3 theta + 3 a cos^2 theta sin theta = n`Prove that `(m + n)^(2/3) + (m - n)^(2/3)`
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
Prove that:
tan (55° + x) = cot (35° – x)
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
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)
Choose the correct alternative:
`(1 + cot^2"A")/(1 + tan^2"A")` = ?
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A
If cosec A – sin A = p and sec A – cos A = q, then prove that `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
If tan θ – sin2θ = cos2θ, then show that sin2 θ = `1/2`.
