Advertisements
Advertisements
प्रश्न
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Advertisements
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 the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
Write the value of `4 tan^2 theta - 4/ cos^2 theta`
Simplify : 2 sin30 + 3 tan45.
If a cos θ − b sin θ = c, then a sin θ + b cos θ =
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1
Choose the correct alternative:
cot θ . tan θ = ?
