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:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
Prove the following trigonometric identities.
`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`
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`
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
\[\frac{1 + \tan^2 A}{1 + \cot^2 A}\]is equal to
Prove the following identity :
`sec^2A + cosec^2A = sec^2Acosec^2A`
Prove the following identity :
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
