Advertisements
Advertisements
प्रश्न
If sinA + cosA = `sqrt(2)` , prove that sinAcosA = `1/2`
Advertisements
उत्तर
We Know , `(sinA + cosA)^2 = sin^2A + cos^2A + 2sinA.cosA`
Given , (sinA + cosA) = `sqrt(2)`
⇒ 2 = 1 + 2sinA.cosA
⇒ 2sinA.cosA = 1
⇒ sinA.cosA = `1/2`
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
If `sin theta = 1/2 , " write the value of" ( 3 cot^2 theta + 3).`
If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
Prove the following identity :
`1/(tanA + cotA) = sinAcosA`
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
If cosθ = `5/13`, then find sinθ.
If `sec θ = 41/40`, then find values of sin θ, cot θ, cosec θ.
Prove that `(1 + sin θ)/(1 - sin θ) = (sec θ + tan θ)^2`.
Prove the following:
(sin α + cos α)(tan α + cot α) = sec α + cosec α
