Question

How do you solve this?
\[\dfrac{1-t{{g}^{2}}{{22.5}^{\circ }}}{tg{{22.5}^{\circ }}}\]

Answer 1

Hint: In the above question, we have been given an expression in the terms of the tangent function. The tangent function has the argument of ${{22.5}^{\circ }}$. For solving it, we can multiply and divide the given expression by two so that the given expression will become \[\dfrac{2\left( 1-{{\tan }^{2}}{{22.5}^{\circ }} \right)}{2\tan {{22.5}^{\circ }}}\]. Then we need to use the trigonometric identity given by $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$ and take its reciprocal to obtain $\dfrac{1-{{\tan }^{2}}x}{2\tan x}=\cot 2x$. Then on substituting $x={{22.5}^{\circ }}$ we will get \[\dfrac{1-{{\tan }^{2}}{{22.5}^{\circ }}}{2\tan {{22.5}^{\circ }}}=\cot {{45}^{\circ }}\] which can be substituted in the expression \[\dfrac{2\left( 1-{{\tan }^{2}}{{22.5}^{\circ }} \right)}{2\tan {{22.5}^{\circ }}}\] to finally obtain its value.

Complete step-by-step solution:
The trigonometric expression given in the above question is
$\Rightarrow \dfrac{1-{{\tan }^{2}}{{22.5}^{\circ }}}{\tan {{22.5}^{\circ }}}$
Multiplying the numerator and the denominator of the above expression by two, we get
\[\begin{align}
  & \Rightarrow \dfrac{2\left( 1-{{\tan }^{2}}{{22.5}^{\circ }} \right)}{2\tan {{22.5}^{\circ }}} \\
 & \Rightarrow 2\left( \dfrac{1-{{\tan }^{2}}{{22.5}^{\circ }}}{2\tan {{22.5}^{\circ }}} \right).......\left( i \right) \\
\end{align}\]
Now, we know the trigonometric identity given by
$\begin{align}
  & \Rightarrow \tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x} \\
 & \Rightarrow \dfrac{2\tan x}{1-{{\tan }^{2}}x}=\tan 2x \\
\end{align}$
Taking reciprocal on both the sides of the above identity we get
$\begin{align}
  & \Rightarrow \dfrac{1-{{\tan }^{2}}x}{2\tan x}=\dfrac{1}{\tan 2x} \\
 & \Rightarrow \dfrac{1-{{\tan }^{2}}x}{2\tan x}=\cot 2x \\
\end{align}$
On substituting $x={{22.5}^{\circ }}$ in the above identity we get
\[\begin{align}
  & \Rightarrow \dfrac{1-{{\tan }^{2}}{{22.5}^{\circ }}}{2\tan {{22.5}^{\circ }}}=\cot 2\left( {{22.5}^{\circ }} \right) \\
 & \Rightarrow \dfrac{1-{{\tan }^{2}}{{22.5}^{\circ }}}{2\tan {{22.5}^{\circ }}}=\cot {{45}^{\circ }} \\
\end{align}\]
Now, we know that $\cot {{45}^{\circ }}=1$. Putting this above we get
\[\Rightarrow \dfrac{1-{{\tan }^{2}}{{22.5}^{\circ }}}{2\tan {{22.5}^{\circ }}}=1\]
Putting the above identity in the (i) expression, we get
$\begin{align}
  & \Rightarrow 2\left( 1 \right) \\
 & \Rightarrow 2 \\
\end{align}$
Hence, the value of the given trigonometric expression is finally found to be equal to $2$.

Note: We have to remember the important trigonometric identities for solving these types of questions. But since the application of the trigonometric identity $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$ was not direct in this question, we have to practice the questions related to the application of the trigonometric identities to quickly get the idea.