Advertisements
Advertisements
प्रश्न
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ −\[\sqrt{3} \tan 3\theta\] is equal to
पर्याय
1
0
−1
\[1 + \sqrt{3}\]
उत्तर
We are given that 5θ and 4θ are acute angles satisfying the following condition sin 5θ = cos 4θ. We are asked to find 2 `sin 3θ -sqrt3 tan 3θ `
⇒ `sin 5θ= cos 4θ`
⇒` cos (90°-5θ)= cos 4θ`
⇒` 90°-5θ=4θ`
⇒ `90=90°`
Where `5θ` and `4θ` are acute angles
⇒ `θ=10°`
Now we have to find:
`2 sin 3θ-sqrt3 tan 3θ`
=` 2 sin 30°-sqrt3 tan 30°`
= `2xx1/2-sqrt3xx1/sqrt3`
=`1-1`
=` 0`
APPEARS IN
संबंधित प्रश्न
If tan A = cot B, prove that A + B = 90
What is the value of (cos2 67° – sin2 23°)?
Solve.
`cos22/sin68`
solve.
sec2 18° - cot2 72°
Evaluate:
3cos80° cosec10° + 2 sin59° sec31°
Use tables to find the acute angle θ, if the value of cos θ is 0.9848
Use tables to find the acute angle θ, if the value of cos θ is 0.9574
Use tables to find the acute angle θ, if the value of tan θ is 0.2419
Prove that:
sec (70° – θ) = cosec (20° + θ)
If A + B = 90° and \[\cos B = \frac{3}{5}\] what is the value of sin A?
If 5 tan θ − 4 = 0, then the value of \[\frac{5 \sin \theta - 4 \cos \theta}{5 \sin \theta + 4 \cos \theta}\] is:
If \[\frac{x {cosec}^2 30°\sec^2 45°}{8 \cos^2 45° \sin^2 60°} = \tan^2 60° - \tan^2 30°\]
The value of cos 1° cos 2° cos 3° ..... cos 180° is
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
Find the sine ratio of θ in standard position whose terminal arm passes through (4,3)
Prove that:
(sin θ + 1 + cos θ) (sin θ − 1 + cos θ) . sec θ cosec θ = 2
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)`
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° - 3A).cosec 42° = 1.
Solve: 2cos2θ + sin θ - 2 = 0.
Find the value of the following:
`cot theta/(tan(90^circ - theta)) + (cos(90^circ - theta) tantheta sec(90^circ - theta))/(sin(90^circ - theta)cot(90^circ - theta)"cosec"(90^circ - theta))`
