Advertisements
Advertisements
प्रश्न
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Advertisements
उत्तर
LHS = `1/(sin θ + cos θ) + 1/(sin θ - cos θ)`
= `((sin θ - cos θ) + (sin θ + cos θ))/(sin^2 θ - cos^2 θ)`
= `(2 sin θ)/((1 - cos^2 θ) - cos^2 θ)`
= `(2 sin θ)/(1 - 2cos^2 θ)`
= RHS
Hence proved.
संबंधित प्रश्न
Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
Show that none of the following is an identity:
`tan^2 theta + sin theta = cos^2 theta`
Write the value of `(1 - cos^2 theta ) cosec^2 theta`.
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
Write the value of tan10° tan 20° tan 70° tan 80° .
\[\frac{x^2 - 1}{2x}\] is equal to
If sin θ = `1/2`, then find the value of θ.
Prove that `"cosec" θ xx sqrt(1 - cos^2θ) = 1`.
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
