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
`(1 + cot^2 theta ) sin^2 theta =1`
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
Write the value of `cosec^2 theta (1+ cos theta ) (1- cos theta).`
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
If sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is ______.
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
