Advertisements
Advertisements
Question
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
Advertisements
Solution
LHS = cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1
= cot θ. cot θ - cosec θ. cosec θ + 1
= (cot2θ - cosec2θ) + 1
= - 1 + 1 = 0
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following identities:
`(i) 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1 = 0`
`(ii) (sin^8 θ – cos^8 θ) = (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ)`
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
`1/(secA + tanA) = secA - tanA`
Prove that:
`tanA/(1 - cotA) + cotA/(1 - tanA) = secA "cosec" A + 1`
Prove that `( sintheta - 2 sin ^3 theta ) = ( 2 cos ^3 theta - cos theta) tan theta`
`If sin theta = cos( theta - 45° ),where theta " is acute, find the value of "theta` .
Prove the following identity :
`1/(tanA + cotA) = sinAcosA`
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
