Advertisements
Advertisements
प्रश्न
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
Advertisements
उत्तर
`tan ((B + C)/2) = cot "A/2`
We know that for a triangle ΔABC
`<A + <B + <C = 180^circ`
`<B + <C = 180^circ - <A`
`(<B + <C)/2 = 90^circ - (<A)/2`
`tan ((B + C)/2) = tan(90^circ - A/2)`
= `cot(A/2)`
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
Prove the following identities:
`(sinAtanA)/(1 - cosA) = 1 + secA`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
Prove that:
2 sin2 A + cos4 A = 1 + sin4 A
If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
Prove the following identity :
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
Evaluate:
`(tan 65^circ)/(cot 25^circ)`
sin(45° + θ) – cos(45° – θ) is equal to ______.
