Question

Evaluate tan 15° + 2 sin 120°:​

Answer 1

$\textbf{To find:}$

$\text{The value of}\;tan\,15^{\circ}+2\;sin\,120^{\circ}$

$\textbf{Solution:}$

$\bf\,tan\,15^{\circ}$

$=tan(45^{\circ}-30^{\circ})$

$=\dfrac{tan\,45^{\circ}-tan\,30^{\circ}}{1+tan\,45^{\circ}\;tan\,30^{\circ}}$

$=\dfrac{1-\frac{1}{\sqrt3}}{1+(1)\frac{1}{\sqrt3}}$

$=\dfrac{\frac{\sqrt{3}-1}{\sqrt3}}{\frac{\sqrt{3}+1}{\sqrt3}}$

$=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

$\implies\boxed{\bf\,tan\,15^{\circ}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}}$

$\bf\,sin\,120^{\circ}$

$=sin(90^{\circ}+30^{\circ})$

$=cos\,30^{\circ}$

$=\dfrac{\sqrt{3}}{2}$

$\implies\boxed{\bf\,sin\,120^{\circ}=\dfrac{\sqrt{3}}{2}}$

$\text{Now,}$

$tan\,15^{\circ}+2\;sin\,120^{\circ}$

$=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+2(\dfrac{\sqrt{3}}{2}})$

$=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\sqrt{3}$

$=\dfrac{\sqrt{3}-1+3+\sqrt{3}}{\sqrt{3}+1}$

$=\dfrac{2\sqrt{3}+2}{\sqrt{3}+1}$

$=\dfrac{2(\sqrt{3}+1)}{\sqrt{3}+1}$

$=2$

$\therefore\textbf{The value of \bf\,tan\,15^{\circ}+2\;sin\,120^{\circ} is 2}$

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