Advertisements
Advertisements
Question
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Advertisements
Solution
LHS = cosec2 (90° - θ) + cot2 (90° - θ)
= sec2 θ + tan2θ
= 1 + tan2θ + tan2θ
= 1 + 2 tan2θ
= RHS
Hence proved.
RELATED QUESTIONS
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
Given `cos38^circ sec(90^circ - 2A) = 1` , Find the value of <A
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
If 1 + sin2α = 3 sinα cosα, then values of cot α are ______.
If sin A = `1/2`, then the value of sec A is ______.
