Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cosA + sinA)^2 + (cosA - sinA)^2 = 2`
Advertisements
उत्तर
LHS = `(cosA + sinA)^2 + (cosA - sinA)^2`
= `cos^2A + sin^2A + 2cosA.sinA + cos^2A + sin^2A - 2cosA.sinA`
= `2(cos^2A + sin^2A) = 2` = RHS
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`
If sin θ + cos θ = x, prove that `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, show that `x^2/a^2 + y^2/b^2 - z^2/c^2 = 1`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
Prove that:
`sqrt(sec^2A + cosec^2A) = tanA + cotA`
If (tan θ + sin θ) = m and (tan θ – sin θ) = n, prove that (m2 – n2)2 = 16 mn.
If `( sin theta + cos theta ) = sqrt(2) , " prove that " cot theta = ( sqrt(2)+1)`.
If cos A + cos2 A = 1, then sin2 A + sin4 A =
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
