Question

Prove that $\tan {15^ \circ } + \cot {15^ \circ } = 4$.

Answer 1

Hint: In order to find the value of $\tan {15^ \circ } + \cot {15^ \circ } = 4$, convert $\tan {15^ \circ }$ in terms of $\cot {15^ \circ }$, as we know that $\tan {15^ \circ } = \dfrac{1}{{\cot {{15}^ \circ }}}$, then multiply and divide the second operand by $\cot {15^ \circ }$, in order to have a common denominator, then solve the numerator above and similarly, applying trigonometric identities solve it further.

Formula used:
$\tan x = \dfrac{1}{{\cot x}}$
\[\sin {30^ \circ } = \dfrac{1}{2}\]
\[2\sin x\cos x = \sin 2x\]
\[\cot x = \dfrac{{\cos x}}{{\sin x}}\]
\[\cos ecx = \dfrac{1}{{\sin x}}\]
\[1 + {\cot ^2}x = \cos e{c^2}x\]

Complete step by step solution:
We are given to prove $\tan {15^ \circ } + \cot {15^ \circ } = 4$.
We will start with the Left-hand side equation, then solving it further to prove it as equal to the right-hand side.
Taking the Left-hand side equation, $\tan {15^ \circ } + \cot {15^ \circ }$.
From trigonometric identities, we know that $\tan x = \dfrac{1}{{\cot x}}$, so applying this in the above equation, we can write it as:
\[\tan {15^ \circ } + \cot {15^ \circ } = \dfrac{1}{{\cot {{15}^ \circ }}} + \cot {15^ \circ }\]
Multiplying and dividing the second operand with $\cot {15^ \circ }$, in order to have a common denominator for both the operands, and we get:
\[
  \dfrac{1}{{\cot {{15}^ \circ }}} + \cot {15^ \circ } \\
   \Rightarrow \dfrac{1}{{\cot {{15}^ \circ }}} + \cot {15^ \circ }\dfrac{{\cot {{15}^ \circ }}}{{\cot {{15}^ \circ }}} \\
   \Rightarrow \dfrac{1}{{\cot {{15}^ \circ }}} + \dfrac{{{{\cot }^2}{{15}^ \circ }}}{{\cot {{15}^ \circ }}} \\
 \]
Since, the denominator is same, so taking the denominator common:
\[\dfrac{{1 + {{\cot }^2}{{15}^ \circ }}}{{\cot {{15}^ \circ }}}\]
From the trigonometric identities, we know that \[1 + {\cot ^2}x = \cos e{c^2}x\].
So, substituting \[1 + {\cot ^2}{15^ \circ } = \cos e{c^2}{15^ \circ }\] in the above equation, we get:
\[\dfrac{{1 + {{\cot }^2}{{15}^ \circ }}}{{\cot {{15}^ \circ }}} = \dfrac{{\cos e{c^2}{{15}^ \circ }}}{{\cot {{15}^ \circ }}}\]

From, some more trigonometric identities, we know that \[\cot x = \dfrac{{\cos x}}{{\sin x}}\] and \[\cos ecx = \dfrac{1}{{\sin x}}\].
So, applying these identities above and we get:
\[
  \dfrac{{\cos e{c^2}{{15}^ \circ }}}{{\cot {{15}^ \circ }}} \\
   \Rightarrow \dfrac{{{{\left( {\dfrac{1}{{\sin {{15}^ \circ }}}} \right)}^2}}}{{\dfrac{{\cos {{15}^ \circ }}}{{\sin {{15}^ \circ }}}}} \\
   \Rightarrow \dfrac{{\dfrac{1}{{{{\sin }^2}{{15}^ \circ }}}}}{{\dfrac{{\cos {{15}^ \circ }}}{{\sin {{15}^ \circ }}}}} \\
 \]
Multiplying the numerator and denominator with \[{\sin ^2}{15^ \circ }\]:
\[
  \dfrac{{\dfrac{1}{{{{\sin }^2}{{15}^ \circ }}}}}{{\dfrac{{\cos {{15}^ \circ }}}{{\sin {{15}^ \circ }}}}} \\
   \Rightarrow \dfrac{{\dfrac{1}{{{{\sin }^2}{{15}^ \circ }}} \times {{\sin }^2}{{15}^ \circ }}}{{\dfrac{{\cos {{15}^ \circ }}}{{\sin {{15}^ \circ }}} \times {{\sin }^2}{{15}^ \circ }}} \\
   \Rightarrow \dfrac{1}{{\cos {{15}^ \circ }\sin {{15}^ \circ }}} \\
 \]

Multiplying the numerator and denominator with \[2\], and we get:
\[\dfrac{1}{{\cos {{15}^ \circ }\sin {{15}^ \circ }}} = \dfrac{2}{{2\sin {{15}^ \circ }\cos {{15}^ \circ }}}\]

From trigonometric angles and sub-angles, we know that \[2\sin x\cos x = \sin 2x\].
Applying this identity in the above equation, and we get:
\[\dfrac{2}{{2\sin {{15}^ \circ }\cos {{15}^ \circ }}} = \dfrac{2}{{\sin 2\left( {{{15}^ \circ }} \right)}} = \dfrac{2}{{\sin {{30}^ \circ }}}\]
Now, as we know that the value of \[\sin {30^ \circ } = \dfrac{1}{2}\].
So, substituting \[\sin {30^ \circ } = \dfrac{1}{2}\]in the above equation, and we get:
\[\dfrac{2}{{\sin {{30}^ \circ }}} = \dfrac{2}{{\dfrac{1}{2}}}\]
Multiplying the numerator and denominator by \[2\] in order to cancel out the denominator, and we get:
\[
  \dfrac{2}{{\sin {{30}^ \circ }}} = \dfrac{2}{{\dfrac{1}{2}}} \\
   \Rightarrow \dfrac{{2 \times 2}}{{\dfrac{1}{2} \times 2}} = \dfrac{4}{1} = 4 \\
   \Rightarrow \dfrac{2}{{\sin {{30}^ \circ }}} = 4 \\
 \]
Which is equal to the Right- hand side value, that means $\tan {15^ \circ } + \cot {15^ \circ } = 4$.

Hence, we prove that $\tan {15^ \circ } + \cot {15^ \circ } = 4$.

Note:
> It’s important to remember the trigonometric formulas to solve this kind of trigonometric equations.
> We can directly solve the equation by putting the value of \[\tan {15^ \circ } = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\] and \[\cot {15^ \circ } = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\] in the equation and then solving it further.