Advertisements
Advertisements
Question
Find A, if 0° ≤ A ≤ 90° and 4 sin2 A – 3 = 0
Advertisements
Solution
4 sin2 A – 3 = 0
`=> sin^2A = 3/4`
`=> sin A = sqrt(3)/2`
We know `sin 60^circ = sqrt(3)/2`
Hence, A = 60°
APPEARS IN
RELATED QUESTIONS
Evaluate:
`cos70^circ/(sin20^circ) + cos59^circ/(sin31^circ) - 8 sin^2 30^circ`
Prove that:
sin (28° + A) = cos (62° – A)
Prove that:
`1/(1 + cos(90^@ - A)) + 1/(1 - cos(90^@ - A)) = 2cosec^2(90^@ - A)`
If \[\tan \theta = \frac{4}{5}\] find the value of \[\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\]
If \[\cos \theta = \frac{2}{3}\] find the value of \[\frac{\sec \theta - 1}{\sec \theta + 1}\]
If \[\tan \theta = \frac{3}{4}\] then cos2 θ − sin2 θ =
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to
Evaluate: `(cos55°)/(sin 35°) + (cot 35°)/(tan 55°)`
If sin 3A = cos 6A, then ∠A = ?
If sin A = `3/5`, then show that 4 tan A + 3 sin A = 6 cos A.
