Advertisements
Advertisements
Question
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
cos(90° - A) · sec 77° = 1
Advertisements
Solution
cos (90° - A) · sec 77° = 1
⇒ cos(90° - A) = `1/(sec 77°)`
⇒ cos(90° - A) = cos 77°
⇒ 90° - A = 77°
⇒ - A = 77° - 90°
⇒ - A = - 13°
⇒ A = 13°
APPEARS IN
RELATED QUESTIONS
`(\text{i})\text{ }\frac{\cot 54^\text{o}}{\tan36^\text{o}}+\frac{\tan 20^\text{o}}{\cot 70^\text{o}}-2`
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solve.
sin42° sin48° - cos42° cos48°
For triangle ABC, show that : `sin (A + B)/2 = cos C/2`
Find the value of x, if tan x = `(tan60^circ - tan30^circ)/(1 + tan60^circ tan30^circ)`
Use tables to find the acute angle θ, if the value of cos θ is 0.6885
Write the acute angle θ satisfying \[\cos B = \frac{3}{5}\]
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
Prove that:
(sin θ + 1 + cos θ) (sin θ − 1 + cos θ) . sec θ cosec θ = 2
Find the value of the following:
`cot theta/(tan(90^circ - theta)) + (cos(90^circ - theta) tantheta sec(90^circ - theta))/(sin(90^circ - theta)cot(90^circ - theta)"cosec"(90^circ - theta))`
