Advertisements
Advertisements
प्रश्न
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
उत्तर
`sin((A + B)/2) = cos"C/2`
We know that for a triangle ΔABC
`<A + <B + <C = 180^circ`
`(<B + <A)/2 = 90^circ - (<C)/2`
`sin((A+B)/2) = sin(90^circ - C/2)`
= `cos(C/2)`
APPEARS IN
संबंधित प्रश्न
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
Prove that:
`1/(sinA - cosA) - 1/(sinA + cosA) = (2cosA)/(2sin^2A - 1)`
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
Prove the following identity :
(secA - cosA)(secA + cosA) = `sin^2A + tan^2A`
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
If tan θ × A = sin θ, then A = ?
Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
