Advertisements
Advertisements
प्रश्न
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
Advertisements
उत्तर
We know that for a triangle ΔABC
∠A + ∠B + ∠C = 180°
∠B + ∠C = 180° – ∠A
`=> (angle B + angle C)/2 = 90^circ - (angle A)/2`
`=> tan ((B + C)/2) = tan (90^circ - A/2)`
= `cot (A/2)`
APPEARS IN
संबंधित प्रश्न
Show that cos 38° cos 52° − sin 38° sin 52° = 0
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
Use tables to find the acute angle θ, if the value of sin θ is 0.4848
Prove that:
tan (55° - A) - cot (35° + A)
The value of tan 10° tan 15° tan 75° tan 80° is
The value of
If θ and 2θ − 45° are acute angles such that sin θ = cos (2θ − 45°), then tan θ is equal to
If ∆ABC is right angled at C, then the value of cos (A + B) is ______.
Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.
Evaluate:
3 cos 80° cosec 10°+ 2 sin 59° sec 31°
