Advertisements
Advertisements
प्रश्न
Prove that `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ)) = 1`.
Advertisements
उत्तर
LHS = `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ))`
= `(sin θ. sin θ cos θ)/(cos θ) + (cos θ . cos θ sin θ)/(sin θ)`
= sin2 θ + cos2 θ
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
`1/((1+ sin θ)) + 1/((1 - sin θ)) = 2 sec^2 θ`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Write the value of `(1 + tan^2 theta ) cos^2 theta`.
\[\frac{x^2 - 1}{2x}\] is equal to
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Prove the following identity :
cosecθ(1 + cosθ)(cosecθ - cotθ) = 1
If x sin3θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ , then show that x2 + y2 = 1.
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A
