Advertisements
Advertisements
प्रश्न
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
Advertisements
उत्तर
LHS = `(sec θ - 1)/(sec θ + 1)`
= `(1/cos θ - 1)/(1/cos θ + 1)`
= `(1 - cos θ)/(1 + cos θ)`
= `(1 - cos θ xx ( 1 + cos θ))/(1 + cos θ xx (1 + cos θ))`
= `(1 - cos^2 θ)/(1 + cos θ)^2`
= `(sin^2 θ)/(1 + cos θ)^2`
= `((sin θ)/(1 + cos θ ))^2`
= RHS
संबंधित प्रश्न
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
`(sec^2 theta -1)(cosec^2 theta - 1)=1`
`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta`
Prove the following identities:
`(sec theta + tan theta)/(sec theta - tan theta) = (sec theta + tan theta)^2 = 1 + 2 tan^2 theta + 2 sec theta tan theta`
What is the value of (1 + cot2 θ) sin2 θ?
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
Without using trigonometric table , evaluate :
`cos90^circ + sin30^circ tan45^circ cos^2 45^circ`
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ
