Advertisements
Advertisements
प्रश्न
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)`
उत्तर
Given: ∠ A = 90°,
∠A + ∠B + ∠C = 180°
∠B + ∠C = 180° − ∠A
`(angle"B"+angle"C")/2 =90° - (angle"A")/2`
⇒ `tan ((angle"B"+angle"C")/2) = tan(90° - (angle"A")/2)`
⇒ `tan ((angle"B"+angle"C")/2) = cot ((angle"A")/2)`
⇒ `tan ((angle"B"+angle"C")/2) = cot ((90°)/2) = cot 45°`
⇒ `tan ((angle"B"+angle"C")/2) = 1`
APPEARS IN
संबंधित प्रश्न
If tan A = cot B, prove that A + B = 90
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Use trigonometrical tables to find tangent of 37°
Use tables to find the acute angle θ, if the value of tan θ is 0.7391
Evaluate:
`2(tan35^@/cot55^@)^2 + (cot55^@/tan35^@)^2 - 3(sec40^@/(cosec50^@))`
Evaluate:
3 cos 80° cosec 10° + 2 cos 59° cosec 31°
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
If A and B are complementary angles, prove that:
cosec2 A + cosec2 B = cosec2 A cosec2 B
Find A, if 0° ≤ A ≤ 90° and 4 sin2 A – 3 = 0
If \[\cos \theta = \frac{2}{3}\] find the value of \[\frac{\sec \theta - 1}{\sec \theta + 1}\]
If \[\tan \theta = \frac{1}{\sqrt{7}}, \text{ then } \frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta} =\]
The value of tan 10° tan 15° tan 75° tan 80° is
If ∆ABC is right angled at C, then the value of cos (A + B) is ______.
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
Prove that:
\[\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2\]
Find the sine ratio of θ in standard position whose terminal arm passes through (4,3)
2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ) is equal to ______.
Sin 2B = 2 sin B is true when B is equal to ______.
If sec A + tan A = x, then sec A = ______.
