Animated Solution for Mathematics - Trigonometry: The sum of solutions of the equation $\frac{\cos x}{1+\sin x}=|\tan 2x|,x\in (-\frac{\pi }{2},\frac{\pi }{2})-\{\frac{\pi }{4},-\frac{\pi }{4}\}$ is :

The problem asks us to solve the equation:

This expression may appear intimidating due to the rational structure and the absolute value, but it possesses a highly symmetric trigonometric structure.

The Half-Angle Magic

To simplify the left-hand side (LHS), we utilize standard half-angle identities. We express the numerator and denominator as follows:

Factoring the numerator as a difference of squares, we obtain:

Dividing both the numerator and denominator by $\cos \left(\frac{x}{2}\right)$, we arrive at the identity:

The Domain Trap

The domain is given as $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)-\left\{\frac{\pi }{4},-\frac{\pi }{4}\right\}$. Within this interval, the angle $\left(\frac{\pi }{4}-\frac{x}{2}\right)$ lies strictly between $0$ and $\frac{\pi }{2}$.

Since the tangent function is positive in the first quadrant, the LHS is guaranteed to be positive. This allows us to equate the expressions without concern for the sign of the absolute value.

The Modulus and the General Solution

To solve $\tan \left(\frac{\pi }{4}-\frac{x}{2}\right)=|\tan 2x|$, we square both sides to eliminate the modulus:

Using the general solution for ${\tan }^2\theta ={\tan }^2\alpha$, which is $\theta =n\pi \pm \alpha$, we set up two cases:

Case 1: $2x=n\pi +\left(\frac{\pi }{4}-\frac{x}{2}\right)\Rightarrow \frac{5x}{2}=n\pi +\frac{\pi }{4}\Rightarrow x=\frac{2n\pi }{5}+\frac{\pi }{10}$

Case 2: $2x=n\pi -\left(\frac{\pi }{4}-\frac{x}{2}\right)\Rightarrow \frac{3x}{2}=n\pi -\frac{\pi }{4}\Rightarrow x=\frac{2n\pi }{3}-\frac{\pi }{6}$

The Final Tally

We now test integer values for $n$ to ensure the solutions fall within the specified domain:

1. For Case 1: $n=0\Rightarrow x=\frac{\pi }{10}$ (Valid); $n=-1\Rightarrow x=-\frac{3\pi }{10}$ (Valid). 2. For Case 2: $n=0\Rightarrow x=-\frac{\pi }{6}$ (Valid); $n=1\Rightarrow x=\frac{\pi }{2}$ (Excluded).

The valid solutions are $x\in \left\{\frac{\pi }{10},-\frac{3\pi }{10},-\frac{\pi }{6}\right\}$. Summing these values:

The final sum of the solutions is $-\frac{11\pi }{30}$.