Advertisements
Advertisements
प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Prove that `(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Advertisements
उत्तर
L.H.S = `(sin theta-2sin^3theta)/(2cos^3theta -costheta)`
= `(sintheta(1-2sin^2theta))/(costheta(2cos^2theta-1))`
= `(sinthetaxx(1-2sin^2theta))/(costhetaxx{2(1-sin^2theta)-1})`
= `(sin thetaxx(1-2sin^2theta))/(costhetaxx(1-2sin^2theta))`
= `tantheta`
= R.H.S
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
`sin^2 theta + cos^4 theta = cos^2 theta + sin^4 theta`
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
Write the value of cos1° cos 2°........cos180° .
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` ...`[cos theta = 1/sectheta]`
