Advertisements
Advertisements
प्रश्न
Express the following in term of angles between 0° and 45° :
sin 59° + tan 63°
उत्तर
sin 59° + tan 63°
= sin(90° – 31°) + tan(90° – 27°)
= cos 31° + cot 27°
APPEARS IN
संबंधित प्रश्न
If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`
Evaluate `(tan 26^@)/(cot 64^@)`
Evaluate.
`cot54^@/(tan36^@)+tan20^@/(cot70^@)-2`
Evaluate:
`(cot^2 41^circ)/(tan^2 49^circ) - 2 sin^2 75^circ/cos^2 15^circ`
Prove that:
`(cos(90^circ - theta)costheta)/cottheta = 1 - cos^2theta`
Use trigonometrical tables to find tangent of 17° 27'
Use tables to find the acute angle θ, if the value of sin θ is 0.4848
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that cos 3 A = 4 cos3 A – 3 cos A
If A + B = 90° and \[\cos B = \frac{3}{5}\] what is the value of sin A?
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
If \[\tan \theta = \frac{1}{\sqrt{7}}, \text{ then } \frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta} =\]
If \[\tan \theta = \frac{3}{4}\] then cos2 θ − sin2 θ =
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
tan 5° ✕ tan 30° ✕ 4 tan 85° is equal to
Prove that:
(sin θ + 1 + cos θ) (sin θ − 1 + cos θ) . sec θ cosec θ = 2
Evaluate: `2(tan57°)/(cot33°) - (cot70°)/(tan20°) - sqrt(2) cos 45°`
The value of 3 sin 70° sec 20° + 2 sin 49° sec 51° is
The value of cosec(70° + θ) – sec(20° − θ) + tan(65° + θ) – cot(25° − θ) is
If x and y are complementary angles, then ______.
