Advertisements
Advertisements
प्रश्न
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
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
Without using trigonometric tables evaluate the following:
`(i) sin^2 25º + sin^2 65º `
What is the value of (cos2 67° – sin2 23°)?
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
Solve.
`sec75/(cosec15)`
Evaluate:
cosec (65° + A) – sec (25° – A)
Evaluate:
cos 40° cosec 50° + sin 50° sec 40°
Write the value of tan 10° tan 15° tan 75° tan 80°?
If 8 tan x = 15, then sin x − cos x is equal to
Prove that :
tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]
Prove that:
cos15° cos35° cosec55° cos60° cosec75° = \[\frac{1}{2}\]
