Advertisements
Advertisements
प्रश्न
A, B and C are interior angles of a triangle ABC. Show that
sin `(("B"+"C")/2) = cos "A"/2`
उत्तर
We know that for a triangle ABC,
∠A + ∠B + ∠C = 180°
∠B + ∠C = 180° − ∠A
`(angle "B" +angle "C")/2 = 90° - (angle"A")/2`
`((sin "B+C")/2) = sin (90°- "A"/2)`
= cos `("A"/2)`
APPEARS IN
संबंधित प्रश्न
If `cosθ=1/sqrt(2)`, where θ is an acute angle, then find the value of sinθ.
Evaluate `(tan 26^@)/(cot 64^@)`
Evaluate cosec 31° − sec 59°
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
if `cosec A = sqrt2` find the value of `(2 sin^2 A + 3 cot^2 A)/(4(tan^2 A - cos^2 A))`
if `3 cos theta = 1`, find the value of `(6 sin^2 theta + tan^2 theta)/(4 cos theta)`
Solve.
`tan47/cot43`
Evaluate:
`cos70^circ/(sin20^circ) + cos59^circ/(sin31^circ) - 8 sin^2 30^circ`
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Find the value of x, if sin 3x = 2 sin 30° cos 30°
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
Use tables to find cosine of 2° 4’
Write the acute angle θ satisfying \[\cos B = \frac{3}{5}\]
If \[\tan A = \frac{5}{12}\] \[\tan A = \frac{5}{12}\] find the value of (sin A + cos A) sec A.
If θ is an acute angle such that \[\tan^2 \theta = \frac{8}{7}\] then the value of \[\frac{\left( 1 + \sin \theta \right) \left( 1 - \sin \theta \right)}{\left( 1 + \cos \theta \right) \left( 1 - \cos \theta \right)}\]
The value of cos2 17° − sin2 73° is
A, B and C are interior angles of a triangle ABC. Show that
If ∠A = 90°, then find the value of tan`(("B+C")/2)`
Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.
