Advertisements
Advertisements
प्रश्न
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Advertisements
उत्तर
LHS = sin (90° - θ) cos (90° - θ)
LHS = cos θ. sin θ
RHS = tan θ. cos2θ
RHS = `sin θ/cos θ` x cos2θ
RHS = cos θ. sin θ
∴ LHS = RHS
Hence proved.
संबंधित प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
What is the value of (1 + cot2 θ) sin2 θ?
What is the value of 9cot2 θ − 9cosec2 θ?
Prove the following identity :
`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)
If `1 - cos^2θ = 1/4`, then θ = ?
Prove that `(sin^2θ)/(cos θ) + cos θ = sec θ`.
