Advertisements
Advertisements
प्रश्न
Express the following in term of angles between 0° and 45° :
cos 74° + sec 67°
Advertisements
उत्तर
cos 74° + sec 67°
= cos(90° – 16°) + sec(90° – 23°)
= sin 16° + cosec 23°
APPEARS IN
संबंधित प्रश्न
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
Solve.
`cos22/sin68`
Evaluate.
`(2tan53^@)/(cot37^@)-cot80^@/tan10^@`
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A – 1 = 0
Write the maximum and minimum values of sin θ.
If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
In the given figure, if AB = 14 cm, BD = 10 cm and DC = 8 cm, then the value of tan B is ______.
If x and y are complementary angles, then ______.
`tan 47^circ/cot 43^circ` = 1
