Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
Advertisements
उत्तर
`(1 - cos^2θ)sec^2θ = tan^2θ`
Consider L.H.S = `sin^2θ1/cos^2θ`
= `tan^2θ` = RHS
APPEARS IN
संबंधित प्रश्न
Prove that `\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }+\frac{\sin\theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{2}{2\sin^{2}\theta -1}`
If sin θ + cos θ = x, prove that `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
If `1 - cos^2θ = 1/4`, then θ = ?
If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ.
Prove that `(tan(90 - θ) + cot(90 - θ))/("cosec" θ) = sec θ`.
Prove that `(sec A)/(tan A + cot A) = sin A`.
