Question

prove that tan85 1/2° = (1+√2)(√2+√3)

Answer 1

Appropriate Question :-

Prove that :-

$\rm \: \: \: \: \: \: tan82\dfrac{1}{2}\degree =  \: (\sqrt{2} + \sqrt{3}) \: (1 + \sqrt{2})$

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

Consider

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

can be rewritten as

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

$\rm \: =  \: cot 7\dfrac{1}{2}\degree$

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

can be rewritten as

$\rm \: =  \: \dfrac{cos\dfrac{15}{2}\degree}{sin\dfrac{15}{2}\degree}$

$\rm \: =  \: \dfrac{2cos^{2} \dfrac{15}{2}\degree}{2 \: cos\dfrac{15}{2}\degree \: sin\dfrac{15}{2}\degree}$

$\rm \: =  \: \dfrac{1 + cos15\degree }{sin15\degree }$

Thus, we get

$\rm \:  tan82\dfrac{1}{2}\degree \: = \: \dfrac{1 + cos15\degree }{sin15\degree } - - - (1)$

Now, Consider

$\rm \: cos15\degree$

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

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

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

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

Now, Consider

$\rm \: sin15\degree$

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

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

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

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

So, on substituting these values in equation (1), we get

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

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

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

On rationalizing the denominator, we get

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

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

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

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

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

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

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

Hence,

$\rm\implies \:\boxed{\sf{\rm \: \:\: tan82\dfrac{1}{2}\degree =  \: (\sqrt{2} + \sqrt{3}) \: (1 + \sqrt{2}) \: }}$

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Formulae Used :-

$\rm \: tan(90\degree - x) = cotx$

$\rm \: sin2x \: = \: 2 \: sinx \: cosx$

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

$\rm \: cos(x - y) = cosx \: cosy \: + \: sinx \: siny$

$\rm \: sin(x - y) = sinx \: cosy \: - \: cosx \: siny$

Answer 2

Step-by-step explanation:

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