Advertisements
Advertisements
प्रश्न
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Advertisements
उत्तर
LHS = cos θ sin (90° - θ) + sin θ cos (90° - θ)
= cos θ. cos θ + sin θ. sin θ
= cos2θ + sin 2θ
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities:
`cosecA + cotA = 1/(cosecA - cotA)`
The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
If sec θ + tan θ = m, show that `(m^2 - 1)/(m^2 + 1) = sin theta`
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
Choose the correct alternative:
1 + cot2θ = ?
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
