Question

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

Answer 1

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

& \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{aligned}$$ Now, we know the trigonometric identity given by $\begin{aligned} & \Rightarrow \tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x} \\\ & \Rightarrow \dfrac{2\tan x}{1-{{\tan }^{2}}x}=\tan 2x \\\ \end{aligned}$ Taking reciprocal on both the sides of the above identity we get $\begin{aligned} & \Rightarrow \dfrac{1-{{\tan }^{2}}x}{2\tan x}=\dfrac{1}{\tan 2x} \\\ & \Rightarrow \dfrac{1-{{\tan }^{2}}x}{2\tan x}=\cot 2x \\\ \end{aligned}$ On substituting $x={{22.5}^{\circ }}$ in the above identity we get $$\begin{aligned} & \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{aligned}$$ Now, we know that $\cot {{45}^{\circ }}=1$. Putting this above we get $$\Rightarrow \dfrac{1-{{\tan }^{2}}{{22.5}^{\circ }}}{2\tan {{22.5}^{\circ }}}=1$$ Putting the above identity in the (i) expression, we get $\begin{aligned} & \Rightarrow 2\left( 1 \right) \\\ & \Rightarrow 2 \\\ \end{aligned}$ **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.