Question

1. Calculate $\rm{4cos^\circ-\sqrt{3}cot20^\circ }$. 2. Calculate $\rm{cosec10^\circ-\sqrt{3}sec10^\circ }$.

Answer 1

$\bigstar$ Solution 1 :-

1. $\sf{4 cos20^\circ-\sqrt{3}cot20^\circ }$

$\sf{= 4cos20^\circ-\sqrt{3}\dfrac{cos20^\circ}{sin20^\circ} }\\\\\sf{= \dfrac{4cos20^\circ.sin20^\circ-\sqrt{3}cos20^\circ }{sin20^\circ} }\\\\\sf{= \dfrac{2sin40^\circ-2.\frac{\sqrt{3}}{2}cos20^\circ }{sin20^\circ} }\\\\\sf{= \dfrac{2sin40^\circ-2.sin60^\circ.cos20^\circ}{sin20^\circ} }\\\\\sf{= \dfrac{2sin40^\circ-(sin80^\circ+sin40^\circ)}{sin20^\circ} }\\\\\sf{= \dfrac{sin40^\circ-sin80^\circ}{sin20^\circ} }\\\\\sf{= \dfrac{2cos60^\circ.sin(-20^\circ)}{sin20^\circ} }\\\\\sf{=-1}$

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$\bigstar$ Solution 2 :-

2. $\sf{cosec10^\circ-\sqrt{3}sec10^\circ }$

$\displaystyle \sf{= \frac{1}{sin10^\circ}-\frac{\sqrt{3} }{cos10^\circ} }\\\\\displaystyle \sf{= \frac{1.cos10^\circ-\sqrt{3}sin10^\circ }{sin10^\circ.cos10^\circ} }\\\\\displaystyle \sf{= \frac{2[\frac{1}{2} cos10^\circ-\frac{\sqrt{3}}{2} sin10^\circ]}{sin10^\circ.cos10^\circ} }\\\\\displaystyle \sf{= \frac{2[\frac{1}{2} cos10^\circ-\frac{\sqrt{3}}{2} sin10^\circ]\times 2}{sin10^\circ.cos10^\circ\times 2} }$

$\displaystyle \sf{= \frac{4[sin30^\circ.cos10^\circ-cos30^\circ.sin10^\circ] }{sin20^\circ} }\\\\\displaystyle \sf{= 4 \frac{sin(30^\circ-sin10^\circ)}{sin20^\circ} }\\\\\displaystyle \sf{= 4 \frac{sin20^\circ}{sin20^\circ} }\\\\\displaystyle \sf{=4}$

Answer 2

Step-by-step explanation:

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