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
`\text{Evaluate }\frac{\tan 65^\circ }{\cot 25^\circ}`
Show that cos 38° cos 52° − sin 38° sin 52° = 0
Prove the following trigonometric identities.
`((1 + cot^2 theta) tan theta)/sec^2 theta = cot theta`
Find the value of angle A, where 0° ≤ A ≤ 90°.
cos (90° – A) . sec 77° = 1
Use tables to find the acute angle θ, if the value of cos θ is 0.6885
Evaluate:
3 cos 80° cosec 10° + 2 cos 59° cosec 31°
Write the value of tan 10° tan 15° tan 75° tan 80°?
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
The value of cos 1° cos 2° cos 3° ..... cos 180° is
The value of \[\frac{\tan 55°}{\cot 35°}\] + cot 1° cot 2° cot 3° .... cot 90°, is
