Advertisements
Advertisements
प्रश्न
Without using tables evaluate: 3cos 80°. cosec 10° + 2sin 59° sec 31°
उत्तर १
3 cos 80°. cosec 10° + 2 sin 59° sec 31°
= 3cos (90° – 10°) cosec 10° + 2sin (90° – 31°) sec 31°
= 3 sin 10° cosec 10° + 2 cos 31°, sec 31° ....[sin (90° – θ) = cos θ, cos (90° – θ) = sin θ]
= 3 × 1 + 2 × 1 .....[∵ sin θ. Cosec θ = 1, cos θ. sec θ = 1]
= 5
उत्तर २
3 cos 80°. cosec 10° + 2 sin 59° sec 31°
⇒ `3 cos (90° - 10°). 1/("sin" 10°) + 2 sin (90° - 31°). 1/(cos 31°)`
⇒ `(3 "sin" 10°)/("sin" 10°) + (2"cos" 31°)/("cos" 31°)`
⇒ 3 + 2 = 5.
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables, evaluate :
`sin 16^circ/cos 74^circ`
Without using trigonometric tables, prove that:
sin53° cos37° + cos53° sin37° = 1
Prove that:
`cos 80^circ/(sin 10^circ) + cos 59^circ "cosec" 31^circ = 2`
Prove that:
\[\frac{\sin\theta \cos(90^\circ - \theta)\cos\theta}{\sin(90^\circ- \theta)} + \frac{\cos\theta \sin(90^\circ - \theta)\sin\theta}{\cos(90^\circ - \theta)}\]
Prove that:
\[cot\theta \tan\left( 90° - \theta \right) - \sec\left( 90° - \theta \right)cosec\theta + \sqrt{3}\tan12° \tan60° \tan78° = 2\]
If sin 3 A = cos (A − 26°), where 3 A is an acute angle, find the value of A.
From the trigonometric table, write the values of tan 45°48'.
Using trigonometric table evaluate the following:
tan 25°45' + cot 45°25'.
`(sin 40° + cos 50°)/(tan 38°20')`
If sin θ = 1, then the value of `1/2 sin(theta/2)`is ______.
