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It is a trigonometry related sum and you have to find the value of the given expression


2 years ago

## Answers : (2)

Arun
24742 Points
							sqrt (3)*csc20° - sec20° = sqrt (3)*(1/sin20°) - 1/cos20° = (sqrt (3)*cos20° - sin20°)/(sin20°*cos20°) = 4*[(sqrt (3)/2)*cos20° - (1/2) sin20°]/(2sin20°cos20°) = 4*(cos30°cos20° - sin30°sin20°)/sin (2*20°) = 4*cos (30°+20°)/sin40° = 4*cos50°/sin (90°-50°) = 4*cos50°/cos50° = 4.

2 years ago
Soumendu Majumdar
159 Points
							Dear Student,Though I know that your question is wrong as per your class since you have posted it I’ll find the value of the expression and you should also know the question should be $\sqrt3 cosec20^{\circ}-sec20^{\circ}$Now to find the value of the expression you have posted we need to know the values of $cosec20^{\circ}$ & $sec20^{\circ}$$cosec20^{\circ} =2.92380440294$(You can use an approximated value)$sec20^{\circ} =1.06417777336$(You can use an approximated value)so,$cosec^220^{\circ} =8.54863218665$$sec^220^{\circ} =1.13247433331$so the value of $\sqrt3 cosec20^{\circ}-sec20^{\circ}$$=13.6741909491$ (Take approximate value)Now to find the value of the correct expression,First, multiply with ½ to both numerator & denominatorso we get$\sqrt3(0.5)cosec20^{\circ}-0.5sec20^\circ$= $4(cos30^{\circ}cos20{^\circ}-sin30^{\circ}sin20{^\circ})/(2sin20{^\circ}cos20{^\circ})$=$4 cos(30+20)^\circ /sin(40)^\circ$=$4 cos50^\circ/sin(90-50)^\circ$=$4 cos50^\circ/cos(50)^\circ$=$4$Hope it helps!

2 years ago
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• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions