Advertisements
Advertisements
Question
Find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° – 3A) . cosec 42° = 1
Advertisements
Solution
sin (90° – 3A) . cosec 42° = 1
`cos3A. 1/(sin42^circ) = 1`
cos 3A = sin 42°
= sin (90° – 48°)
= cos 48°
3A = 48°
A = 16°
APPEARS IN
RELATED QUESTIONS
If tan A = cot B, prove that A + B = 90°.
Evaluate `(tan 26^@)/(cot 64^@)`
Evaluate:
`3 sin72^circ/(cos18^circ) - sec32^circ/(cosec58^circ)`
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
Prove that:
tan (55° - A) - cot (35° + A)
Prove that:
sec (70° – θ) = cosec (20° + θ)
Find A, if 0° ≤ A ≤ 90° and 4 sin2 A – 3 = 0
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
If sec A + tan A = x, then sec A = ______.
