Advertisements
Advertisements
प्रश्न
For triangle ABC, show that : `sin (A + B)/2 = cos C/2`
उत्तर
We know that for a triangle ΔABC
∠A + ∠B + ∠C = 180°
`(angleB + angleA)/2 = 90^circ - (angleC)/2`
`sin((A + B)/2) = sin(90^circ - C/2)`
= `cos(C/2)`
APPEARS IN
संबंधित प्रश्न
`\text{Evaluate }\frac{\tan 65^\circ }{\cot 25^\circ}`
If tan A = cot B, prove that A + B = 90
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
Evaluate:
`(5sin66^@)/(cos24^@) - (2cot85^@)/(tan5^@)`
If sin θ =7/25, where θ is an acute angle, find the value of cos θ.
Find the value of the following:
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin31^circ) + cos theta/(sin(90^circ - theta))- 8cos^2 60^circ`
`(sin 75^circ)/(cos 15^circ)` = ?
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
In ∆ABC, cos C = `12/13` and BC = 24, then AC = ?
If y sin 45° cos 45° = tan2 45° – cos2 30°, then y = ______.
