Advertisements
Advertisements
प्रश्न
Solve: 2cos2θ + sin θ - 2 = 0.
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°
APPEARS IN
संबंधित प्रश्न
For triangle ABC, show that : `sin (A + B)/2 = cos C/2`
Evaluate:
14 sin 30° + 6 cos 60° – 5 tan 45°
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Prove that:
`1/(1 + sin(90^@ - A)) + 1/(1 - sin(90^@ - A)) = 2sec^2(90^@ - A)`
If tanθ = 2, find the values of other trigonometric ratios.
What is the maximum value of \[\frac{1}{\sec \theta}\]
If θ is an acute angle such that \[\tan^2 \theta = \frac{8}{7}\] then the value of \[\frac{\left( 1 + \sin \theta \right) \left( 1 - \sin \theta \right)}{\left( 1 + \cos \theta \right) \left( 1 - \cos \theta \right)}\]
Prove that:
(sin θ + 1 + cos θ) (sin θ − 1 + cos θ) . sec θ cosec θ = 2
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° - 3A).cosec 42° = 1.
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
