Advertisements
Advertisements
प्रश्न
Evaluate:
14 sin 30° + 6 cos 60° – 5 tan 45°
Advertisements
उत्तर
14 sin 30° + 6 cos 60° – 5 tan 45°
= `14(1/2) + 6(1/2) - 5(1)`
= 7 + 3 – 5
= 5
संबंधित प्रश्न
Show that : `sin26^circ/sec64^circ + cos26^circ/(cosec64^circ) = 1`
Evaluate:
`2 tan57^circ/(cot33^circ) - cot70^circ/(tan20^circ) - sqrt(2) cos45^circ`
Prove that:
sin (28° + A) = cos (62° – A)
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
Write the maximum and minimum values of cos θ.
If \[\frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta}\] write the value of \[\frac{1 - \cos^2 \theta}{2 - \sin^2 \theta}\]
If tan2 45° − cos2 30° = x sin 45° cos 45°, then x =
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
Prove that:
\[\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2\]
If A, B and C are interior angles of a ΔABC then `cos (("B + C")/2)` is equal to ______.
