find the value of 1/cos290+1/(3)^1/2sin 250is equal to
find the value of 1/cos290+1/(3)^1/2sin 250is equal to
Munish , 7 Years ago
Grade 11
Trigonometry> find the value of 1/cos290+1/(3)^1/2sin 2...
1 Answers
Arun
Last Activity: 7 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
LIVE ONLINE CLASSES
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Other Related Questions on trigonometry
In ABC, if cot A+cot B+cot C = cosecA cosecB cosecC, then prove that ABC is right angle triangle.
In ABC, if cot A+cot B+cot C = cosecA cosecB cosecC, then prove that ABC is right angle triangle.
trigonometry
1 Answer Available
Last Activity: 2 Years ago
- Find the general solution of equation tan3∂=cot2∂
For all angles between -720 and +720
- Find the general solution of equation tan3∂=cot2∂
For all angles between -720 and +720
trigonometry
0 Answer Available
Last Activity: 3 Years ago
To prove : Cosec(45 degree - A)Cosec(45 degree + A) = 2SecA
To prove : Cosec(45 degree - A)Cosec(45 degree + A) = 2SecA
trigonometry
0 Answer Available
Last Activity: 3 Years ago
cos^2 13ºcos^2 47º+cos^2 73º =? …..................
cos^2 13ºcos^2 47º+cos^2 73º =? …..................
trigonometry
0 Answer Available
Last Activity: 3 Years ago
Derive a formula for the area of the triangle, given b, A, B, and C. That is, derive the formula: (b^2sinAsinC)/(2sinB)
Derive a formula for the area of the triangle, given b, A, B, and C. That is, derive the formula: (b^2sinAsinC)/(2sinB)
trigonometry
0 Answer Available
Last Activity: 3 Years ago