Advertisements
Advertisements
प्रश्न
Find the value of the following:
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin31^circ) + cos theta/(sin(90^circ - theta))- 8cos^2 60^circ`
Advertisements
उत्तर
cos 60° = `1/sqrt(2)`
`(cos 70^circ)/(sin 20^circ) = (cos(90^circ - 20^circ))/(sin 20^circ) = (sin 20^circ)/(sin 20^circ)` = 1
`(cos 59^circ)/(sin 31^circ) = (cos(90^circ - 31^circ))/(sin 31^circ) = (sin 31^circ)/(sin 31^circ)` = 1
`(cos theta)/(sin(90^circ - theta)) = cos theta/cos theta` = 1
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin31^circ) + cos theta/(sin(90^circ - theta))- 8cos^2 60^circ`
= `1 + 1 + 1 - 8(1/2)^2`
= `3 - 8 xx 1/4`
= 3 – 2
= 1
APPEARS IN
संबंधित प्रश्न
Evaluate:
`sin80^circ/(cos10^circ) + sin59^circ sec31^circ`
Use trigonometrical tables to find tangent of 42° 18'
Evaluate:
`(5sin66^@)/(cos24^@) - (2cot85^@)/(tan5^@)`
If θ is an acute angle such that \[\tan^2 \theta = \frac{8}{7}\] then the value of \[\frac{\left( 1 + \sin \theta \right) \left( 1 - \sin \theta \right)}{\left( 1 + \cos \theta \right) \left( 1 - \cos \theta \right)}\]
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to
Prove that:
cos15° cos35° cosec55° cos60° cosec75° = \[\frac{1}{2}\]
Prove that:
(sin θ + 1 + cos θ) (sin θ − 1 + cos θ) . sec θ cosec θ = 2
Evaluate: 14 sin 30°+ 6 cos 60°- 5 tan 45°.
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
cos(90° - A) · sec 77° = 1
If sec A + tan A = x, then sec A = ______.
