Advertisements
Advertisements
Question
Evaluate:
`3 sin72^circ/(cos18^circ) - sec32^circ/(cosec58^circ)`
Advertisements
Solution
`3 sin72^circ/(cos18^circ) - sec32^circ/(cosec58^circ)`
= `3 sin(90^circ - 18^circ)/(cos18^circ) - sec(90^circ - 58^circ)/(cosec58^circ)`
= `3 cos18^circ/(cos18^circ) - (cosec58^circ)/(cosec58^circ)`
= 3 – 1
= 2
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
Evaluate:
3cos80° cosec10° + 2 sin59° sec31°
Evaluate:
`cos70^circ/(sin20^circ) + cos59^circ/(sin31^circ) - 8 sin^2 30^circ`
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
If 3 cot θ = 4, find the value of \[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\]
If \[\tan A = \frac{3}{4} \text{ and } A + B = 90°\] then what is the value of cot B?
If \[\frac{x {cosec}^2 30°\sec^2 45°}{8 \cos^2 45° \sin^2 60°} = \tan^2 60° - \tan^2 30°\]
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to
If cot (90 – A) = 1, then ∠A = ?
