Advertisements
Advertisements
प्रश्न
Evaluate: 14 sin 30°+ 6 cos 60°- 5 tan 45°.
उत्तर
14 sin 30°+ 6 cos 60°- 5 tan 45°
= `14(1/2) + 6(1/2) - 5(1)`
= 7 + 3 - 5
= 5.
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables, evaluate the following:
`( i)\frac{\cos37^\text{o}}{\sin53^\text{o}}\text{ }(ii)\frac{\sin41^\text{o}}{\cos 49^\text{o}}(iii)\frac{\sin30^\text{o}17'}{\cos59^\text{o}\43'}`
Show that : sin 42° sec 48° + cos 42° cosec 48° = 2
Evaluate:
cos 40° cosec 50° + sin 50° sec 40°
Evaluate:
`(3sin72^@)/(cos18^@) - sec32^@/(cosec58^@)`
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A + cos A – 1 = 0
If \[\tan \theta = \frac{4}{5}\] find the value of \[\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\]
The value of \[\frac{\cos^3 20°- \cos^3 70°}{\sin^3 70° - \sin^3 20°}\]
Evaluate: `(cos55°)/(sin 35°) + (cot 35°)/(tan 55°)`
The value of tan 1° tan 2° tan 3°…. tan 89° is
`tan 47^circ/cot 43^circ` = 1
