Question no. 6 pure.Value of trigonometric function
Tatv Shah , 6 Years ago
Grade 11
Trigonometry> Question no. 6 pure. Value of trigonometr...
1 Answers
Arun
Last Activity: 6 Years ago
cos 290° = cos (270°+20°) = sin 20°
. . . . . . . . sin 250° = sin (270°-20°) = - cos 20°.
∴ ( 1 / cos 290° ) + (1 / √3 sin 250° )
= ( 1 / sin 20° ) + { 1 / [ √3 ( - cos 20° ) ] }
= ( √3 cos 20° - sin 20° ) / ( √3 sin 20° cos 20° )
= 2 [ (√3 /2 ) cos 20° - (1/2) sin 20° ] / [ (√3/2) ( 2 sin 40°]
= 2 ( sin 60° cos 20° - cos 60° sin 20° ) / ((√3/2) sin 40° )
= 2 sin (60° - 20°) / ((√3 /2) sin 40° )
= 4 / √3
. . . . . . . . sin 250° = sin (270°-20°) = - cos 20°.
∴ ( 1 / cos 290° ) + (1 / √3 sin 250° )
= ( 1 / sin 20° ) + { 1 / [ √3 ( - cos 20° ) ] }
= ( √3 cos 20° - sin 20° ) / ( √3 sin 20° cos 20° )
= 2 [ (√3 /2 ) cos 20° - (1/2) sin 20° ] / [ (√3/2) ( 2 sin 40°]
= 2 ( sin 60° cos 20° - cos 60° sin 20° ) / ((√3/2) sin 40° )
= 2 sin (60° - 20°) / ((√3 /2) sin 40° )
= 4 / √3
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