Question

What is the expression $\dfrac{1-{{\tan }^{2}}\left( {{45}^{\circ }}-A \right)}{1+{{\tan }^{2}}\left( {{45}^{\circ }}-A \right)}$ is equal to
A. $\sin 2A$
B. $\cos 2A$
C. $\tan 2A$
D. $\cot 2A$

Answer 1

At first, we replace ${{\tan }^{2}}\left( {{45}^{\circ }}-A \right)$ with $\dfrac{{{\sin }^{2}}\left( {{45}^{\circ }}-A \right)}{{{\cos }^{2}}\left( {{45}^{\circ }}-A \right)}$ . Then, we multiply the numerator and denominator by ${{\cos }^{2}}\left( {{45}^{\circ }}-A \right)$ and get $\dfrac{{{\cos }^{2}}\left( {{45}^{\circ }}-A \right)-{{\sin }^{2}}\left( {{45}^{\circ }}-A \right)}{{{\cos }^{2}}\left( {{45}^{\circ }}-A \right)+{{\sin }^{2}}\left( {{45}^{\circ }}-A \right)}$ . We then use the formula ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ to replace ${{\cos }^{2}}\left( {{45}^{\circ }}-A \right)+{{\sin }^{2}}\left( {{45}^{\circ }}-A \right)$ by $1$ . We then use the formula ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta$ to replace ${{\cos }^{2}}\left( {{45}^{\circ }}-A \right)-{{\sin }^{2}}\left( {{45}^{\circ }}-A \right)$ by $\cos 2\left( {{45}^{\circ }}-A \right)$ . Rewriting $\cos 2\left( {{45}^{\circ }}-A \right)$ as $\cos \left( {{90}^{\circ }}-2A \right)$ and using the formula $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta$ , we get the final answer.

Complete step by step solution:
The given expression that we have is,
$\dfrac{1-{{\tan }^{2}}\left( {{45}^{\circ }}-A \right)}{1+{{\tan }^{2}}\left( {{45}^{\circ }}-A \right)}$
We know the formula that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . Using this in the above expression, we get,
= \dfrac{1-{{\left\\{ \dfrac{\sin \left( {{45}^{\circ }}-A \right)}{\cos \left( {{45}^{\circ }}-A \right)} \right\\}}^{2}}}{1+{{\left\\{ \dfrac{\sin \left( {{45}^{\circ }}-A \right)}{\cos \left( {{45}^{\circ }}-A \right)} \right\\}}^{2}}}
The above expression can be simplified as,
$= \dfrac{1-\dfrac{{{\sin }^{2}}\left( {{45}^{\circ }}-A \right)}{{{\cos }^{2}}\left( {{45}^{\circ }}-A \right)}}{1+\dfrac{{{\sin }^{2}}\left( {{45}^{\circ }}-A \right)}{{{\cos }^{2}}\left( {{45}^{\circ }}-A \right)}}$
Multiplying the numerator and denominator by ${{\cos }^{2}}\left( {{45}^{\circ }}-A \right)$ , we get,
$= \dfrac{{{\cos }^{2}}\left( {{45}^{\circ }}-A \right)-{{\sin }^{2}}\left( {{45}^{\circ }}-A \right)}{{{\cos }^{2}}\left( {{45}^{\circ }}-A \right)+{{\sin }^{2}}\left( {{45}^{\circ }}-A \right)}$
Now, we know the formula that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ . Using this in the above expression, we get,
$= \dfrac{{{\cos }^{2}}\left( {{45}^{\circ }}-A \right)-{{\sin }^{2}}\left( {{45}^{\circ }}-A \right)}{1}$
Now, we also know the formula that ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta$ . Using this in the above expression, we get,
$= \cos 2\left( {{45}^{\circ }}-A \right)$
Multiplying $2$ inside the bracket, we get,
$= \cos \left( {{90}^{\circ }}-2A \right)$
We know the formula that $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta$ . Using this in the above expression, we get,
$= \sin 2A$
Thus, we can conclude that the value of the expression $\dfrac{1-{{\tan }^{2}}\left( {{45}^{\circ }}-A \right)}{1+{{\tan }^{2}}\left( {{45}^{\circ }}-A \right)}$ is $\sin 2A$ which is option A.

Note: Finding the value of the expression by step-by-step solving is necessary and useful in understanding the concepts and also in subjective questions. But for rapid tests like competitive exams, where only the final answer is required, we can check by putting A as ${{45}^{\circ }}$ . This gives the expression as,
$\Rightarrow \dfrac{1-{{\tan }^{2}}\left( {{45}^{\circ }}-{{45}^{\circ }} \right)}{1+{{\tan }^{2}}\left( {{45}^{\circ }}-{{45}^{\circ }} \right)}=\dfrac{1-0}{1+0}=1$ .
Now, putting A as ${{45}^{\circ }}$ in the first option, we get,
$\sin \left( 2\times {{45}^{\circ }} \right)=\sin {{90}^{\circ }}=1$
Putting A as ${{45}^{\circ }}$ in the other options, we do not get $1$ as the answer. So, the first option is the correct option.