Advertisements
Advertisements
प्रश्न
Find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° – 3A) . cosec 42° = 1
Advertisements
उत्तर
sin (90° – 3A) . cosec 42° = 1
`cos3A. 1/(sin42^circ) = 1`
cos 3A = sin 42°
= sin (90° – 48°)
= cos 48°
3A = 48°
A = 16°
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
(cosecA − sinA) (secA − cosA) (tanA + cotA) = 1
Evaluate:
`(cot^2 41^circ)/(tan^2 49^circ) - 2 sin^2 75^circ/cos^2 15^circ`
Prove that:
`1/(1 + cos(90^@ - A)) + 1/(1 - cos(90^@ - A)) = 2cosec^2(90^@ - A)`
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
If tanθ = 2, find the values of other trigonometric ratios.
If 3 cot θ = 4, find the value of \[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\]
If 16 cot x = 12, then \[\frac{\sin x - \cos x}{\sin x + \cos x}\]
Prove that :
tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° - 3A).cosec 42° = 1.
Find the value of the following:
`((cos 47^circ)/(sin 43^circ))^2 + ((sin 72^circ)/(cos 18^circ))^2 - 2cos^2 45^circ`
