Advertisements
Advertisements
प्रश्न
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
Advertisements
उत्तर
LHS = cosec2(90° - θ) - tan2 θ
LHS = sec2 θ - tan2 θ = 1
RHS = cos2(90° - θ) + cos2 θ
RHS = sin2θ + cos2 θ = 1
Hence, LHS = RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities:
cosec A(1 + cos A) (cosec A – cot A) = 1
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
`(cot ^theta)/((cosec theta+1)) + ((cosec theta + 1))/cot theta = 2 sec theta`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
The value of sin ( \[{45}^° + \theta) - \cos ( {45}^°- \theta)\] is equal to
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
If 2sin2θ – cos2θ = 2, then find the value of θ.
(1 + sin A)(1 – sin A) is equal to ______.
