Advertisements
Advertisements
प्रश्न
Evaluate:
`(cot^2 41^circ)/(tan^2 49^circ) - 2 sin^2 75^circ/cos^2 15^circ`
उत्तर
`(cot^2 41^circ)/(tan^2 49^circ) - 2 sin^2 75^circ/cos^2 15^circ`
= `[cot(90^circ - 49^circ)]^2/(tan^2 49^circ) - 2 [sin(90^circ - 15^circ)]^2/cos^2 15^circ`
= `tan^2 49^circ/(tan^2 49^circ) - 2 cos^2 15^circ/cos^2 15^circ`
= 1 – 2
= – 1
APPEARS IN
संबंधित प्रश्न
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solve.
sin15° cos75° + cos15° sin75°
Use tables to find cosine of 65° 41’
Use trigonometrical tables to find tangent of 37°
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that sin 3 A = 3 sin A – 4 sin3 A
Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
If 16 cot x = 12, then \[\frac{\sin x - \cos x}{\sin x + \cos x}\]
Find the value of the following:
`((cos 47^circ)/(sin 43^circ))^2 + ((sin 72^circ)/(cos 18^circ))^2 - 2cos^2 45^circ`
Find the value of the following:
sin 21° 21′
If tan θ = 1, then sin θ . cos θ = ?
