Advertisements
Advertisements
प्रश्न
Find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° – 3A) . cosec 42° = 1
Advertisements
उत्तर
sin (90° – 3A) . cosec 42° = 1
`cos3A. 1/(sin42^circ) = 1`
cos 3A = sin 42°
= sin (90° – 48°)
= cos 48°
3A = 48°
A = 16°
संबंधित प्रश्न
If tan A = cot B, prove that A + B = 90°.
Evaluate `(tan 26^@)/(cot 64^@)`
Prove the following trigonometric identities.
`((1 + cot^2 theta) tan theta)/sec^2 theta = cot theta`
Prove the following trigonometric identities.
(cosecθ + sinθ) (cosecθ − sinθ) = cot2 θ + cos2θ
Evaluate.
cos225° + cos265° - tan245°
Use trigonometrical tables to find tangent of 42° 18'
Use tables to find the acute angle θ, if the value of sin θ is 0.6525
If 3 cot θ = 4, find the value of \[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\]
The value of cos2 17° − sin2 73° is
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
