Advertisements
Advertisements
प्रश्न
If tanθ = 2, find the values of other trigonometric ratios.
उत्तर
tanθ = 2 ...(Given)
∴ cotθ = `1/2`
We have,
1 + tan2θ = sec2θ
1 + (2)2 = sec2θ
1 + 4 = sec2θ
5 = sec2θ
Taking Square root on both sides
secθ = `sqrt5`
Cosθ = `1/secθ`
Cosθ = `1/sqrt5`
tanθ = `sinθ/cosθ`
2 = `sinθ/(1/sqrt5)`
2 = `sinθ xx sqrt5/1`
`2/sqrt5 = sinθ`
`sinθ = 2/sqrt5`
cosecθ = `1/sinθ`
= `1/(2/sqrt5)`
= `1 xx sqrt5/2`
∴ cosecθ = `sqrt5/2`
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}`
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
if `sqrt3 tan theta = 3 sin theta` find the value of `sin^2 theta - cos^2 theta`
Solve.
`tan47/cot43`
For triangle ABC, show that : `sin (A + B)/2 = cos C/2`
Prove that:
`1/(1 + sin(90^@ - A)) + 1/(1 - sin(90^@ - A)) = 2sec^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 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that sin 3 A = 3 sin A – 4 sin3 A
If 0° < A < 90°; find A, if `sinA/(secA - 1) + sinA/(secA + 1) = 2`
Write the value of tan 10° tan 15° tan 75° tan 80°?
If A + B = 90° and \[\tan A = \frac{3}{4}\]\[\tan A = \frac{3}{4}\] what is cot B?
If \[\frac{x {cosec}^2 30°\sec^2 45°}{8 \cos^2 45° \sin^2 60°} = \tan^2 60° - \tan^2 30°\]
Prove the following.
tan4θ + tan2θ = sec4θ - sec2θ
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
Evaluate:
3 cos 80° cosec 10°+ 2 sin 59° sec 31°
Find the value of the following:
sin 21° 21′
The value of tan 72° tan 18° is
The value of (tan1° tan2° tan3° ... tan89°) is ______.