Question

Simplify the following

$tan82 \frac{1}{2} \degree$

Explain all the steps involved.


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Brainly Star

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Answer 1

$\large\underline{\sf{Solution-}}$

Given Trigonometric expression is

$\rm \: tan82 \frac{1}{2}\degree$

can be rewritten as

$\rm \: = \: tan\bigg(90\degree - 7 \frac{1}{2}\degree \bigg)$

$\rm \: = \: cot7 \frac{1}{2}\degree$

$\rm \: = \: cot\frac{15}{2}\degree$

Let assume that

$\rm \:x = \: \frac{15}{2}\degree$

So,

$\rm \: cotx =\dfrac{cosx}{sinx}$

can be rewritten as

$\rm \: cotx =\dfrac{2cosx \times cosx}{2cosx \times sinx}$

$\rm \: cotx =\dfrac{ {2cos}^{2}x }{sin2x}$

$\rm \: cotx =\dfrac{1 + cos2x}{sin2x}$

On substituting the value of x, we again

$\rm \: cot \frac{15}{2}\degree =\dfrac{1 + cos15\degree}{sin15\degree} - - - (1)$

Now, Consider

$\rm \: cos15\degree = cos(45\degree - 30\degree)$

$\rm \: cos15\degree = cos45\degree \: cos30\degree + sin45\degree \: sin30\degree$

$\rm \: cos15\degree = \dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} + \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{2}$

$\rm\implies \:cos15\degree = \dfrac{ \sqrt{3} + 1}{2 \sqrt{2} } - - - (2)$

Now, Consider

$\rm \: sin15\degree = sin(45\degree - 30\degree)$

$\rm \: sin15\degree = sin45\degree \: cos30\degree \: - \: cos45\degree \: sin30\degree$

$\rm \: sin15\degree = \dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} - \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{2}$

$\rm\implies \:sin15\degree = \dfrac{ \sqrt{3} - 1}{2 \sqrt{3} } - - - (2)$

On substituting the values from equation (2) and (3) in equation (1), we get

$\rm \: cot \frac{15}{2}\degree =\dfrac{1 + cos15\degree}{sin15\degree}$

$\rm \: = \: \dfrac{1 + \dfrac{ \sqrt{3} + 1}{2 \sqrt{2} } }{\dfrac{ \sqrt{3} - 1}{2 \sqrt{2} } }$

$\rm \: = \: \dfrac{\dfrac{2 \sqrt{2} + \sqrt{3} + 1}{2 \sqrt{2} } }{\dfrac{ \sqrt{3} - 1}{2 \sqrt{2} } }$

$\rm \: = \: \dfrac{2 \sqrt{2} + \sqrt{3} + 1}{ \sqrt{3} - 1}$

$\rm \: = \: \dfrac{2 \sqrt{2} + \sqrt{3} + 1}{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1}$

$\rm \: = \: \dfrac{2 \sqrt{6} + 3 + \sqrt{3} + 2 \sqrt{2} + \sqrt{3} + 1}{ (\sqrt{3})^{2} - {1}^{2} }$

$\rm \: = \: \dfrac{2 \sqrt{6} + \sqrt{3} + 2 \sqrt{2} + \sqrt{3} + 4}{ 3 - 1}$

$\rm \: = \: \dfrac{2\sqrt{6} +2 \sqrt{3} + 2 \sqrt{2}+ 4}{ 2}$

$\rm \: = \: \dfrac{2(\sqrt{6} + \sqrt{3} + \sqrt{2}+ 2)}{ 2}$

$\rm \: = \: \sqrt{6} + \sqrt{3} + \sqrt{2} + 2$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \:cot7 {\dfrac{1}{2} }^{\degree} = \: \sqrt{6} + \sqrt{3} + \sqrt{2} + 2 \: \: }}$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:sin2x = 2sinx \: cosx \: }}$

$\boxed{ \rm{ \:cos2x = {2cos}^{2}x - 1 \: }}$

$\boxed{ \rm{ \:cos(x - y) = cosxcosy + sinxsiny \: }}$

$\boxed{ \rm{ \:sin(x - y) = sinxcosy - cosxsiny \: }}$

$\rule{190pt}{2pt}$

$\begin{gathered} { \boxed{ \begin{array}{c} \underline{\underline{ \color{orange} \text{Additional \: lnformation}}} \\& \rm \: sin2x \: = 2 \: sinx \: cosx\:\\ & \rm \: cos2x = 1 - {2sin}^{2}x \\ & \rm \: cos2x = {2cos}^{2}x - 1 \\ & \rm \: cos2x = {cos}^{2}x - {sin}^{2}x \\ & \rm \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} \\ & \rm \: sin2x = \dfrac{2tanx}{1 + {tan}^{2}x } \\ & \rm \:sin3x = 3sinx - {4sin}^{3}x \\ & \rm \: cos3x = {4cos}^{3}x - 3cosx \\ & \rm \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - {3tan}^{2}x} \end{array}}}\end{gathered}$

Answer 2

Step-by-step explanation:

$\begin{array}{l} \rm \displaystyle \tan 30=\frac{1}{\sqrt{3}} \\ \\ \displaystyle\rm \frac{2 \tan 15}{1-\tan ^{2} 15}=\frac{1}{\sqrt{3}} \text{ Let} \tan 15=a \\ \\ \displaystyle\rm 2 \sqrt{3} a=1-a^{2} \\ \\ \displaystyle\rm a^{2}+2 \sqrt{3} a-1=0 \\ \\ \displaystyle\rm a=\frac{-2 \sqrt{3} \pm \sqrt{12+4}}{2} \quad(\because \tan 15>0) \\ \\ \displaystyle\rm \therefore \tan 15=2-\sqrt{3} \\ \\ \displaystyle\rm \displaystyle\rm \frac{2 \tan 7 \dfrac{1}{2}^{\circ}}{1-\tan ^{2} 7 \dfrac{1}{2}^{\circ}}-2-\sqrt{3} \: \: \: \: \:\\ \\ \displaystyle \rm \text{Let} \tan 7 \frac{1}{2} ^{ \circ}=b \\ \\ \displaystyle\rm 2 b=(2-\sqrt{3})\left(1-b^{2}\right) \\ \\ \displaystyle\rm (2-\sqrt{3}) b^{2}+2 b-(2-\sqrt{3})=0 \\ \\ \displaystyle\rm \therefore b=\frac{-2+\sqrt{4+4(2-\sqrt{3})^{2}}}{2(2-\sqrt{3})} \quad\left(\because \tan 7 \frac{1}{2}^{\circ}>0\right) \\ \\ \rm \displaystyle=\frac{-1+\sqrt{8-4 \sqrt{3}}}{(2-\sqrt{3})} \\ \\ \displaystyle\rm \therefore \tan 7 \frac{1}{2}^{\circ}=\frac{\sqrt{6}-\sqrt{2}-1}{2-\sqrt{3}} \\ \\ \displaystyle\rm \tan 82 \frac{1}{2}^{\circ}=\cot 7 \frac{1}{2}^{\circ}=\frac{2-\sqrt{3}}{\sqrt{6}-\sqrt{2}-1} \end{array}$

$\begin{array}{l} \displaystyle \rm=\frac{(2-\sqrt{3})(\sqrt{6}+\sqrt{2}+1)}{(\sqrt{6}-(\sqrt{2}+1))(\sqrt{6}+\sqrt{2}+1)} \\ \\ \displaystyle \rm =\frac{2 \sqrt{6}-3 \sqrt{2}+2 \sqrt{2}-\sqrt{6}+2-\sqrt{3}}{6-(\sqrt{2}+1)^{2}} \\ \\ \displaystyle \rm =\frac{\sqrt{6}-\sqrt{2}+2-\sqrt{3}}{3-2 \sqrt{2}} \\ \\ \displaystyle \rm =\frac{(\sqrt{2}-1)(\sqrt{3}+\sqrt{2})}{(\sqrt{2}-1)^{2}} \\ \\ \displaystyle \rm =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}-1}=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1) \\ \\ \color{olive} \boxed {\displaystyle \rm \therefore \tan 82 \frac{1}{2}^{ \degree} =(\sqrt{3}+\sqrt{2})(\sqrt{2}+1) }\end{array}$