Question

tan165degree ki value ​

Answer 1

Answer:

$\tan \left(165^{\circ \:}\right)=\sqrt{3}-2\quad \begin{pmatrix}\mathrm{Decimal:}&-0.26794\dots \end{pmatrix}$

Step-by-step explanation:

$\tan \left(165^{\circ \:}\right)$

$\gray{\mathrm{Write}\:\tan \left(165^{\circ \:}\right)\:\mathrm{as}\:\tan \left(30^{\circ \:}+135^{\circ \:}\right)}$

$=\tan \left(30^{\circ \:}+135^{\circ \:}\right)$

$\displaystyle\gray{\mathrm{Using\:the\:summation\:identity}:\quad \tan \left(x+y\right)=\frac{\tan \left(x\right)+\tan \left(y\right)}{1-\tan \left(x\right)\tan \left(y\right)}}$

$=\displaystyle\frac{\tan \left(30^{\circ \:}\right)+\tan \left(135^{\circ \:}\right)}{1-\tan \left(30^{\circ \:}\right)\tan \left(135^{\circ \:}\right)}$

$\displaystyle\gray{\mathrm{Use\:the\:following\:trivial\:identity}:\quad \tan \left(30^{\circ \:}\right)=\frac{\sqrt{3}}{3}}$

$\black{\tan \left(135^{\circ \:}\right)}$

$\displaystyle\gray{\mathrm{Use\:the\:following\:identity}:\quad \tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}}$

$\displaystyle=\frac{\sin \left(135^{\circ \:}\right)}{\cos \left(135^{\circ \:}\right)}$

$\displaystyle\gray{\sin \left(135^{\circ \:}\right)=\frac{\sqrt{2}}{2}}\\\gray{\cos \left(135^{\circ \:}\right)=-\frac{\sqrt{2}}{2}}$

$\displaystyle=\frac{\frac{\sqrt{2}}{2}}{\left(-\frac{\sqrt{2}}{2}\right)}$

$\displaystyle\frac{\displaystyle\frac{\sqrt{2}}{2}}{\left(-\displaystyle\frac{\sqrt{2}}{2}\right)}$

$\displaystyle\gray{\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{-b}=-\frac{a}{b}}$

$\displaystyle=-\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$\displaystyle\gray{\mathrm{Divide\:fractions}:\quad \frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot \:d}{b\cdot \:c}}$

$\displaystyle=-\frac{\sqrt{2}\cdot \:2}{2\sqrt{2}}$

$\gray{\mathrm{Cancel\:the\:common\:factor:}\:\sqrt{2}}$

$\displaystyle=-\frac{2}{2}$

$\gray{\mathrm{Cancel\:the\:common\:factor:}\:2}$

$=-1$

$\displaystyle=\frac{\displaystyle\frac{\sqrt{3}}{3}+\left(-1\right)}{1-\displaystyle\frac{\sqrt{3}}{3}\left(-1\right)}$

$\displaystyle\black{\frac{\frac{\sqrt{3}}{3}+\left(-1\right)}{1-\frac{\sqrt{3}}{3}\left(-1\right)}}$

$\gray{\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a,\:-\left(-a\right)=a}$

$\displaystyle=\frac{\frac{\sqrt{3}}{3}-1}{1+\frac{\sqrt{3}}{3}\cdot \:1}$

$\displaystyle\gray{\mathrm{Multiply:}\:\frac{\sqrt{3}}{3}\cdot \:1=\frac{\sqrt{3}}{3}}$

$\displaystyle=\frac{\frac{\sqrt{3}}{3}-1}{1+\frac{\sqrt{3}}{3}}$

$\displaystyle\gray{\mathrm{Join}\:1+\frac{\sqrt{3}}{3}:\quad \frac{\sqrt{3}+1}{\sqrt{3}}}$

$\displaystyle=\frac{\frac{\sqrt{3}}{3}-1}{\frac{\sqrt{3}+1}{\sqrt{3}}}$

$\displaystyle\gray{\mathrm{Join}\:\frac{\sqrt{3}}{3}-1:\quad \frac{1-\sqrt{3}}{\sqrt{3}}}$

$\displaystyle=\frac{\frac{1-\sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$

$\displaystyle\gray{\mathrm{Divide\:fractions}:\quad \frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot \:d}{b\cdot \:c}}$

$\displaystyle=\frac{\left(1-\sqrt{3}\right)\sqrt{3}}{\sqrt{3}\left(\sqrt{3}+1\right)}$

$\gray{\mathrm{Cancel\:the\:common\:factor:}\:\sqrt{3}}$

$\displaystyle=\frac{1-\sqrt{3}}{\sqrt{3}+1}$

$\displaystyle\gray{\mathrm{Rationalize\:}\frac{1-\sqrt{3}}{\sqrt{3}+1}:\quad \sqrt{3}-2}$

$=\sqrt{3}-2$