Advertisements
Advertisements
Question
Without using tables evaluate: 3cos 80°. cosec 10° + 2sin 59° sec 31°
Solution 1
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
Solution 2
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
RELATED QUESTIONS
In the below given figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20√3 m long. Find the sun’s altitude.
Without using trigonometric tables, evaluate :
`sin 16^circ/cos 74^circ`
Without using trigonometric tables, evaluate :
`tan 27^circ/cot 63^circ`
Without using trigonometric tables, prove that:
sin53° cos37° + cos53° sin37° = 1
Prove that:
`(sin 70^circ)/(cos 20^circ) + ("cosec" 20^circ)/(sec 70^circ) - 2 cos 70^circ "cosec" 20^circ = 0`
Prove that:
sin θ cos (90° - θ ) + sin (90° - θ) cos θ = 1
Prove that:
\[\frac{\cos(90^\circ - \theta)}{1 + \sin(90^\circ - \theta)} + \frac{1 + \sin(90^\circ- \theta)}{\cos(90^\circ - \theta)} = 2 cosec\theta\]
Prove that:
cos1° cos2° cos3° ... cos180° = 0
Prove the following:
`1/(1+sin^2theta) + 1/(1+cos^2theta) + 1/(1+sec^2theta) + 1/(1+cosec^2theta) = 2`
Using trigonometric table evaluate the following:
cos 64°42' - sin 42°20'
