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Grade 11Trigonometry

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

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Aryan Kumar

8 Years agoGrade 11

2 Answers

Arun

8 Years ago

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.

Soumendu Majumdar

8 Years ago

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 & denominator

so 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!

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