Question

Show that :
$\rm \: tan75 \degree = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } = 2 + \sqrt{3}$
Hence deduce that
$\rm \: tan \: 75 \degree - cot 75 \degree = 4sin60 \degree$
Relevant Answers Needed !​

Answer 1

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

Consider,

$\rm \: tan75\degree$

$\rm \: = \: tan(45\degree + 30\degree )$

We know,

$\boxed{\tt{ tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany}}}$

So, using this identity, we get

$\rm \: = \: \dfrac{tan45\degree + tan30\degree }{1 - tan45\degree \: tan30\degree }$

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

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

So, on rationalizing the denominator, we get

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

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

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

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

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

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

Hence,

$\rm\implies \:\boxed{\tt{ tan75\degree = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } = 2 + \sqrt{3} \: }}$

Now, Consider

$\rm \: tan75\degree + cot75\degree$

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

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

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

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

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

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

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

can be rewritten as

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

$\rm \: = \: 4 sin60\degree$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \:tan75\degree - cot75\degree = \: 4 sin60\degree \: }}$

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Basic Identities Used

$\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: }}$

$\boxed{\tt{ {x}^{2} - {y}^{2} = (x + y)(x - y) \: }}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ sin(x + y) = sinxcosy + sinycosx \: }}$

$\boxed{\tt{ sin(x - y) = sinxcosy - sinycosx \: }}$

$\boxed{\tt{ cos(x + y) = cosx \: cosy \: - \: sinx \: siny \: }}$

$\boxed{\tt{ cos(x - y) = cosx \: cosy \: + \: sinx \: siny \: }}$

$\boxed{\tt{ tan(x - y) = \frac{tanx - tany}{1 + tanx \: tany} \: }}$

$\boxed{\tt{ {sin}^{2}x - {sin}^{2}y = sin(x + y)sin(x - y) \: }}$

$\boxed{\tt{ {cos}^{2}x - {sin}^{2}y = cos(x + y)cos(x - y) \: }}$

Answer 2

$\mid\overline{\underline{\bf Question}} \mid$

Show that

$\rm \: tan75 \degree = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } = 2 + \sqrt{3}$

Hence deduce that

$\rm \: tan \: 75 \degree - cot 75 \degree = 4sin60 \degree$

$\huge\tt Solution$

Simply while proving the above trigonometrical identity first of all we'll start with LHS i.e, tan 75° and bring it equal to 2 + ✓3 after that we will bring the middle term to the same condition by simply rationalizing it.

In second part , in place of tan 75° we'll put the above value and hence prove the second equation.

Now , Let's Proceed !

$\rm \: tan75 \degree = \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } = 2 + \sqrt{3}$

Take L.H.S (or First term)

$\tan75 \degree = \tan(45 \degree+ 30 \degree)$

We break it to 45° and 30° because their value are known !

We know that ,

$\underbrace{ \overbrace{ \rm \tan(A + B) = \dfrac{tanA + tanB}{1 - tanA tanB} }}$

Therefore,

$\rm \tan(45 \degree + 30 \degree) = \dfrac{tan45 \degree + tan30 \degree}{1 - tan45 \degree \: tan30 \degree}$

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

Now , observe that we have prove first term equal to the second term now let's prove these both terms equal to the third term .

On rationalizing the (2) eq

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

$\rm\dfrac{ (\sqrt{3}) ^{2} + 1 + 2 \sqrt{3} }{3 - 1} = \dfrac{3 + 1 + 2 \sqrt{3} }{2} \\ = \frac{4 + 2 \sqrt{3} }{2} = \frac{ \cancel2(2 + \sqrt{3})}{ \cancel2} \\ \red{= 2 + \sqrt{3} }$

Hence , proved the above trigonometrical identity.

Part 2

Now , instead of tan 75 ° we'll use the above obtained value.

Before that Note the below procedure for cot 75 °

$\cot 75 ° = \dfrac{1}{ \tan75 °} \\ = \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} } = \dfrac{2 - \sqrt{3} }{4 - 3} \\ \red{ = 2 - \sqrt{3} } \longrightarrow( \star)$

Now use this in other proving Identity

$= 2 + \sqrt{3} - (2 - \sqrt{3}) \\ = \cancel2 + \sqrt{3} - \cancel2 - \sqrt{3} \\ = 2 \sqrt{3}$

Multiply and divide it with 2 .

$2 \sqrt{3} = 4 \times \dfrac{ \sqrt{3} }{2} \\ \red{= 4 \sin60 \degree }$

Hence , Deduced ✔︎

Thankyou