Advertisements
Advertisements
Question
Evaluate:
3cos80° cosec10° + 2 sin59° sec31°
Advertisements
Solution
sin (90° = θ) = cos θ
cos (90° = θ) = sinθ
3 sin (90° - 80°) cosec 10° + 2 cos (90° - 59°) sec 31°
3 sin 10° cosec 10° + 2 cos 31° sec31°
sin θ cosecθ = 1, cos θ secθ = 1
3 + 2 = 5
RELATED QUESTIONS
Without using trigonometric tables, evaluate the following:
`(\sin ^{2}20^\text{o}+\sin^{2}70^\text{o})/(\cos ^{2}20^\text{o}+\cos ^{2}70^\text{o}}+\frac{\sin (90^\text{o}-\theta )\sin \theta }{\tan \theta }+\frac{\cos (90^\text{o}-\theta )\cos \theta }{\cot \theta }`
Without using trigonometric tables evaluate:
`(sin 65^@)/(cos 25^@) + (cos 32^@)/(sin 58^@) - sin 28^2. sec 62^@ + cosec^2 30^@`
If `tan theta = 1/sqrt2` find the value of `(cosec^2 theta - sec^2 theta)/(cosec^2 theta + cot^2 theta)`
If `tan theta = 12/5` find the value of `(1 + sin theta)/(1 -sin theta)`
Use tables to find sine of 21°
Prove that:
tan (55° - A) - cot (35° + A)
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
If sin θ =7/25, where θ is an acute angle, find the value of cos θ.
The value of tan 1° tan 2° tan 3°…. tan 89° is
In ∆ABC, `sqrt(2)`AC = BC, sin A = 1, sin2A + sin2B + sin2C = 2, then ∠A = ?, ∠B = ?, ∠C = ?
