Question

Prove that $\tan 75 + \cot 75 = 4$.

Answer 1

To prove that $\tan 75 + \cot 75 = 4$, first of all we will be writing 75 as 45 plus 30.
So, we can write $\tan 75$ as $\tan \left( {45 + 30} \right)$. Now, we have the formula for
$\Rightarrow \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Using these, we will find the value of $\tan 75$ and then we know that cot is inverse of tan. Hence, we will get the value of $\cot 75$ as well. Now, we simply need to add these two values and check if the answer is 4 or not.

Complete step-by-step answer:
In this question, we have to prove that $\tan 75 + \cot 75 = 4$.
For proving this, we need to use some trigonometric relations and formulas and also some mathematical calculations.
Now, we do not have any direct formula for proving this. So, we need to use some mathematical calculations.
First of all we can write 75 as 45 plus 30. Therefore,
$\tan 75 = \tan \left( {45 + 30} \right)$
And $\cot 75 = \cot \left( {45 + 30} \right)$
Now, let us find the values of $\tan 75$ and $\cot 75$ separately.
$\Rightarrow \tan \left( {45 + 30} \right)$
Now, we have a formula
$\Rightarrow \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Here, A is 45 and B is 30. Therefore,
$\Rightarrow \tan \left( {45 + 30} \right) = \dfrac{{\tan 45 + \tan 30}}{{1 - \tan 45\tan 30}}$
Now, $\tan 45 = 1$ and $\tan 30 = \dfrac{1}{{\sqrt 3 }}$. Therefore,
$\Rightarrow \tan \left( {45 + 30} \right) = \dfrac{{1 + \dfrac{1}{{\sqrt 3 }}}}{{1 - \left( 1 \right)\left( {\dfrac{1}{{\sqrt 3 }}} \right)}} \\\ \Rightarrow \tan \left( {45 + 30} \right) = \dfrac{{\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}}} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}} \\\$
And, for $\cot 75$, we know that cot is reciprocal of tan. So therefore,
$\Rightarrow \cot 75 = \dfrac{1}{{\tan 75}}$
And we have found the value of $\tan 75 = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}$.
Therefore,
$\Rightarrow \cot 75 = \dfrac{1}{{\tan 75}} = \dfrac{1}{{\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}}} = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$
Now, we have both the values and now we need to add them. Therefore,
$\Rightarrow \tan 75 + \cot 75 = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}} + \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$
Now, taking LCM, we get

$$\Rightarrow \tan 75 + \cot 75 = \dfrac{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 + 1} \right) + \left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} \\\ \Rightarrow \tan 75 + \cot 75 = \dfrac{{3 + \sqrt 3 + \sqrt 3 + 1 + 3 - \sqrt 3 - \sqrt 3 + 1}}{{3 + \sqrt 3 - \sqrt 3 - 1}} \\\ \Rightarrow \tan 75 + \cot 75 = \dfrac{{8 + 2\sqrt 3 - 2\sqrt 3 }}{2} \\\ \Rightarrow \tan 75 + \cot 75 = \dfrac{8}{2} \\\ \Rightarrow \tan 75 + \cot 75 = 4 \\\$$

Hence, LHS = RHS.
Therefore, we have proved $\tan 75 + \cot 75 = 4$.

Note: We can also use another method for proving that $\tan 75 + \cot 75 = 4$.
First of all, we know that $\cot 75 = \dfrac{1}{{\tan 75}}$. Therefore, we get
$\Rightarrow \tan 75 + \dfrac{1}{{\tan 75}} = 4$- - - - - - - (1)
Taking LCM, we get
$\Rightarrow \dfrac{{{{\tan }^2}75 + 1}}{{\tan 75}} = 4 \\\ \Rightarrow {\tan ^2}75 + 1 = 4\tan 75 \\\ \Rightarrow {\tan ^2}75 - 4\tan 75 + 1 = 0 \\\$
Now, let tan75 be equal to x.
$\tan 75 = x$
Therefore,
$\Rightarrow {x^2} - 4x + 1 = 0$
Using the quadratic formula, we get
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}} = \dfrac{{4 \pm \sqrt {12} }}{2}$
$\Rightarrow x = \dfrac{{4 + \sqrt {12} }}{2} = \dfrac{{4 + 2\sqrt 3 }}{2} = 2 + \sqrt 3$
Therefore, $\tan 75 = 2 + \sqrt 3$
Put this value in equation (1), we get
$\Rightarrow \tan 75 + \dfrac{1}{{\tan 75}} = 4 \\\ \Rightarrow \left( {2 + \sqrt 3 } \right) + \dfrac{1}{{\left( {2 + \sqrt 3 } \right)}} = 4 \\\ \Rightarrow \dfrac{{\left( {2 + \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right) + 1}}{{\left( {2 + \sqrt 3 } \right)}} = 4 \\\ \Rightarrow \dfrac{{8 + 4\sqrt 3 }}{{2 + \sqrt 3 }} = 4 \\\ \Rightarrow \dfrac{{8 + 4\sqrt 3 }}{{2 + \sqrt 3 }} \times \dfrac{{\left( {2 - \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)}} = 4 \\\ \Rightarrow \dfrac{{16 - 8\sqrt 3 + 8\sqrt 3 - 12}}{{4 - 2\sqrt 3 + 2\sqrt 3 - 3}} = 4 \\\ \Rightarrow 4 = 4 \\\$
Hence, we have proved that $\tan 75 + \cot 75 = 4$.