Advertisements
Advertisements
Question
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
Advertisements
Solution
L.H.S. = `tan^2A - tan^2B`
= `sin^2A/cos^2A - sin^2B/cos^2B`
= `(sin^2A * cos^2B - sin^2B * cos^2A)/(cos^2A * cos^2B`
= `(sin^2A(1 - sin^2B)-sin^2B(1 - sin^2A))/(cos^2A * cos^2B)`
= `(sin^2A - sin^2A * sin^2B - sin^2B + sin^2A * sin^2B)/(cos^2A * cos^2B`
= `(sin^2A - cancel(sin^2A * sin^2B) - sin^2B + cancel(sin^2A * sin^2B))/(cos^2A * cos^2B`
= `(sin^2A - sin^2B)/(cos^2A * cos^2B)` = R.H.S.
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
`1/(1 + sin A) + 1/(1 - sin A) = 2sec^2 A`
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
If `( tan theta + sin theta ) = m and ( tan theta - sin theta ) = n " prove that "(m^2-n^2)^2 = 16 mn .`
Prove the following identity :
`sqrt((1 + cosA)/(1 - cosA)) = cosecA + cotA`
If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` ...`[sin^2"A" + square = 1]`
= `square` – cos2A ...[sin2A = 1 – cos2A]
= `square`
= R.H.S.
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
