Advertisements
Advertisements
प्रश्न
Solve: 2cos2θ + sin θ - 2 = 0.
Advertisements
उत्तर
2cos2θ + sin θ - 2 = 0
⇒ 2( 1 - sin2θ) + sin θ - 2 = 0
⇒ 2 - 2 sin2θ + sin θ - 2 = 0
⇒ - sin θ( 2 sin θ - 1) = 0
⇒ sin θ( 2 sin θ - 1) = 0
⇒ sin θ = 0 or 2 sin θ - 1 = 0
⇒ sin θ = 0 or sin θ = `1/2`
⇒ θ = 30°
संबंधित प्रश्न
Without using trigonometric tables, evaluate the following:
`(\sin ^{2}20^\text{o}+\sin^{2}70^\text{o})/(\cos ^{2}20^\text{o}+\cos ^{2}70^\text{o}}+\frac{\sin (90^\text{o}-\theta )\sin \theta }{\tan \theta }+\frac{\cos (90^\text{o}-\theta )\cos \theta }{\cot \theta }`
if `cot theta = 1/sqrt3` find the value of `(1 - cos^2 theta)/(2 - sin^2 theta)`
Use tables to find the acute angle θ, if the value of sin θ is 0.6525
If A + B = 90° and \[\cos B = \frac{3}{5}\] what is the value of sin A?
In the following Figure. AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ.
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
A, B and C are interior angles of a triangle ABC. Show that
sin `(("B"+"C")/2) = cos "A"/2`
The value of cosec(70° + θ) – sec(20° − θ) + tan(65° + θ) – cot(25° − θ) is
Prove that `(tan A)/(cot A) = (sec^2A)/("cosec"^2A)`.
In the given figure, if AB = 14 cm, BD = 10 cm and DC = 8 cm, then the value of tan B is ______.
