Question

$\tan \left[\left(\frac{\pi }{4}\right)+\frac{1}{2}{\cos }^{-1}\left(x\right)\right]+\tan \left[\left(\frac{\pi }{4}\right)-\frac{1}{2}{\cos }^{-1}\left(x\right)\right]=1$ has how many solutions?

• A. One solution
• B. Two solutions
• C. Three solutions
• D. No solution

Answer 1

The correct option is D

No solution

Explanation for the correct option:
Finding the number of solutions to the expression:

$\tan \left[\left(\frac{\pi }{4}\right)+\frac{1}{2}{\cos }^{-1}\left(x\right)\right]+\tan \left[\left(\frac{\pi }{4}\right)-\frac{1}{2}{\cos }^{-1}\left(x\right)\right]=1$

Step-1 : Assumption

Let us put $\frac{1}{2}{\cos }^{-1}\left(x\right)=a$.

Then $\cos \left(2a\right)=x$.

So, the above expression $\tan \left[\left(\frac{\pi }{4}\right)+\frac{1}{2}{\cos }^{-1}\left(x\right)\right]+\tan \left[\left(\frac{\pi }{4}\right)-\frac{1}{2}{\cos }^{-1}\left(x\right)\right]=1$will reduces to :

$\tan \left[\left(\frac{\pi }{4}\right)+a\right]+\tan \left[\left(\frac{\pi }{4}\right)-a\right]=1$.

Step-2 : Simplification

Formula to be used : We know that, $\tan \left(A+B\right)=\frac{\tan \left(A\right)+\tan \left(B\right)}{1-\tan \left(A\right)\tan \left(B\right)}$ and $\tan \left(A-B\right)=\frac{\tan \left(A\right)-\tan \left(B\right)}{1+\tan \left(A\right)\tan \left(B\right)}$

So,

$$\begin{gathered} \tan \left[\left(\frac{\pi }{4}\right)+a\right]=\frac{1+\tan \left(a\right)}{1-\tan \left({\displaystyle \frac{\pi }{4}}\right)\tan \left(a\right)} \\ =\frac{1+\tan \left(a\right)}{1-\tan \left(a\right)}\left[\because tan\left(\frac{\pi }{4}\right)=1\right] \end{gathered}$$

Similarly,

$$\begin{gathered} \tan \left[\left(\frac{\pi }{4}\right)-a\right]=\frac{1-\tan \left(a\right)}{1+\tan \left({\displaystyle \frac{\pi }{4}}\right)\tan \left(a\right)} \\ =\frac{1-\tan \left(a\right)}{1+\tan \left(a\right)}\left[\because tan\left(\frac{\pi }{4}\right)=1\right] \end{gathered}$$

So the expression will be :

$\frac{1+\tan \left(a\right)}{1-\tan \left(a\right)}+\frac{1-\tan \left(a\right)}{1+\tan \left(a\right)}=1$.

Step-3 : Finding the solution (if exists)

$$\begin{gathered} \frac{1+\tan \left(a\right)}{1-\tan \left(a\right)}+\frac{1-\tan \left(a\right)}{1+\tan \left(a\right)}=1 \\ \Rightarrow \frac{{\left(1+\tan \left(a\right)\right)}^2+{\left(1-\tan \left(a\right)\right)}^2}{\left(1+\tan \left(a\right)\right)\left(1-\tan \left(a\right)\right)}=1 \\ \Rightarrow \frac{2\left(1+{\tan }^2\left(a\right)\right)}{1-{\tan }^2\left(a\right)}=1\mathbf{}\left[\because {\left(1+tan\left(a\right)\right)}^2+{\left(1-tan\left(a\right)\right)}^2=2\left(1+ta{n}^2\left(a\right)\right),\left(1+tan\left(a\right)\right)\left(1-tan\left(a\right)\right)=1-ta{n}^2\left(a\right)\right] \\ \Rightarrow \frac{1-{\tan }^2\left(a\right)}{1+{\tan }^2\left(a\right)}=2 \\ \Rightarrow \cos \left(2a\right)=2\mathbf{}\left[\because \frac{1-ta{n}^2\left(a\right)}{1+ta{n}^2\left(a\right)}=cos\left(2a\right)\right] \end{gathered}$$

Which does not give any value of $a$ because $\cos \left(2a\right)=2$ cannot be possible as, for any real number $\theta$, $-1\le \cos \theta \le 1$.

Therefore, the given equation has no solution.

Hence, option (D) is the correct answer.