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 sin θ =3/5, where θ is an acute angle, find the value of cos θ.
Write all the other trigonometric ratios of ∠A in terms of sec A.
if `sqrt3 tan theta = 3 sin theta` find the value of `sin^2 theta - cos^2 theta`
Solve.
`tan47/cot43`
Evaluate:
`3 sin72^circ/(cos18^circ) - sec32^circ/(cosec58^circ)`
Evaluate:
3cos80° cosec10° + 2 sin59° sec31°
Evaluate:
`2 tan57^circ/(cot33^circ) - cot70^circ/(tan20^circ) - sqrt(2) cos45^circ`
Use tables to find cosine of 8° 12’
Use tables to find the acute angle θ, if the value of sin θ is 0.6525
Use tables to find the acute angle θ, if the value of cos θ is 0.9848
If A and B are complementary angles, prove that:
cosec2 A + cosec2 B = cosec2 A cosec2 B
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A + cos A – 1 = 0
What is the maximum value of \[\frac{1}{\sec \theta}\]
If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]
Without using trigonometric tables, prove that:
sec70° sin20° + cos20° cosec70° = 2
Express the following in term of angles between 0° and 45° :
cos 74° + sec 67°
Evaluate: `2(tan57°)/(cot33°) - (cot70°)/(tan20°) - sqrt(2) cos 45°`
The value of tan 1° tan 2° tan 3°…. tan 89° is
In ∆ABC, `sqrt(2)` AC = BC, sin A = 1, sin2A + sin2B + sin2C = 2, then ∠A = ? , ∠B = ?, ∠C = ?