Maths / Introduction to Trigonometry / Trigonometric ratios for 60 degrees

QUESTION

The value for the expression$\frac{\sqrt{13}\;cos \;60^0+2\;sin\; 60^0}{\sqrt{13}\;cos\; 60^0 -2\;sin\;60^0}$  is ____________.

 OPTIONS A. $\sqrt{13}-2\sqrt3$$\sqrt{13}-2\sqrt3$ B. $\sqrt{13}+2\sqrt3$ C. $(\sqrt{13}-2\sqrt3)^2$ D. $(\sqrt{13}+2\sqrt3)^2$
Right Option : D

EXPLANATION
Explain TypeExplanation Content
Text

$\frac{\sqrt{13}\;cos \;60^0+2\;sin\; 60^0}{\sqrt{13}\;cos\; 60^0 -2\;sin\;60^0}$

$\frac{\sqrt{13}\frac{1}{2}+2\frac{\sqrt3}{2}}{\sqrt{13}\frac{1}{2}- 2\frac{\sqrt3}{2}}= \frac{13+2\sqrt3}{13-2\sqrt3}$

$= \frac{\sqrt{13}+2\sqrt3}{\sqrt{13}-2\sqrt3} X \frac{\sqrt{13}+2\sqrt3}{\sqrt{13}+2\sqrt3}= \frac{(\sqrt{13}+2\sqrt3)^2}{(\sqrt{13})^2-(2\sqrt3)^2}= (\sqrt{13}+2\sqrt3)^2$

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