Advertisements
Advertisements
Question
Find the value of angle A, where 0° ≤ A ≤ 90°.
cos (90° – A) . sec 77° = 1
Advertisements
Solution
cos (90° – A) . sec 77° = 1
`sinA. 1/(cos77^circ) = 1`
sin A = cos 77°
= cos (90° – 13°)
= sin 13°
A = 13°
APPEARS IN
RELATED QUESTIONS
Express each of the following in terms of trigonometric ratios of angles between 0º and 45º;
(i) cosec 69º + cot 69º
(ii) sin 81º + tan 81º
(iii) sin 72º + cot 72º
Evaluate.
`(cos^2 32^@+cos^2 58^@)/(sin^2 59^@+sin^2 31^@)`
Evaluate:
cosec (65° + A) – sec (25° – A)
Find the value of x, if cos (2x – 6) = cos2 30° – cos2 60°
If 3 cos θ = 5 sin θ, then the value of
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
If \[\cos \theta = \frac{2}{3}\] then 2 sec2 θ + 2 tan2 θ − 7 is equal to
Express the following in term of angles between 0° and 45° :
cos 74° + sec 67°
Prove that `(tan A)/(cot A) = (sec^2A)/("cosec"^2A)`.
In the given figure, if AB = 14 cm, BD = 10 cm and DC = 8 cm, then the value of tan B is ______.

