Advertisements
Advertisements
प्रश्न
Express the following in term of angles between 0° and 45° :
sin 59° + tan 63°
Advertisements
उत्तर
sin 59° + tan 63°
= sin(90° – 31°) + tan(90° – 27°)
= cos 31° + cot 27°
संबंधित प्रश्न
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A
Evaluate `(sin 18^@)/(cos 72^@)`
if `3 cos theta = 1`, find the value of `(6 sin^2 theta + tan^2 theta)/(4 cos theta)`
if `sqrt3 tan theta = 3 sin theta` find the value of `sin^2 theta - cos^2 theta`
Evaluate.
`cot54^@/(tan36^@)+tan20^@/(cot70^@)-2`
For triangle ABC, show that : `sin (A + B)/2 = cos C/2`
Evaluate:
`2 tan57^circ/(cot33^circ) - cot70^circ/(tan20^circ) - sqrt(2) cos45^circ`
Find the value of x, if sin 2x = 2 sin 45° cos 45°
Find the value of angle A, where 0° ≤ A ≤ 90°.
cos (90° – A) . sec 77° = 1
Prove that:
`(cos(90^circ - theta)costheta)/cottheta = 1 - cos^2theta`
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
Use tables to find sine of 62° 57'
Evaluate:
`sec26^@ sin64^@ + (cosec33^@)/sec57^@`
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to
Without using trigonometric tables, prove that:
sec70° sin20° + cos20° cosec70° = 2
Evaluate: `(cos55°)/(sin 35°) + (cot 35°)/(tan 55°)`
Evaluate: `(sin 80°)/(cos 10°)`+ sin 59° sec 31°
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
cos(90° - A) · sec 77° = 1
