Advertisements
Advertisements
Question
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
Advertisements
Solution
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)`
RELATED QUESTIONS
`\text{Evaluate }\frac{\tan 65^\circ }{\cot 25^\circ}`
if `sqrt3 tan theta = 3 sin theta` find the value of `sin^2 theta - cos^2 theta`
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A + cos A – 1 = 0
If 0° < A < 90°; find A, if `sinA/(secA - 1) + sinA/(secA + 1) = 2`
If \[\tan A = \frac{3}{4} \text{ and } A + B = 90°\] then what is the value of cot B?
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
Prove that :
tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]
Express the following in term of angles between 0° and 45° :
cosec 68° + cot 72°
Evaluate: 14 sin 30°+ 6 cos 60°- 5 tan 45°.
If y sin 45° cos 45° = tan2 45° – cos2 30°, then y = ______.
