Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

If $\alpha +\beta +\gamma =\pi$ and $\tan \left[\frac{\alpha +\beta -\gamma }{4}\right]\tan \left[\frac{\gamma +\alpha -\beta }{4}\right]\tan \left[\frac{\gamma +\beta -\alpha }{4}\right]=1$ then the value of $1+\cos \alpha +\cos \beta +\cos \gamma$ is $\mathrm{K}-1$ where K is

$$\begin{array}{rlrlr}& \Rightarrow \tan A\tan B\tan C=1& \frac{\sin A\sin B}{\cos A\cos B}=\frac{1}{\tan C}\Rightarrow \frac{\sin A\sin B-\cos A\cos B}{\sin A\sin B+\cos A\cos B}=\frac{1-\tan C}{1+\tan C}& \Rightarrow \frac{-\cos (A+B)}{\cos (A-B)}=\frac{\sin \left(\frac{\pi }{4}-C\right)}{\cos \left(\frac{\pi }{4}-C\right)}& \Rightarrow 2\sin \left(\frac{\pi }{4}-C\right)\cos (A-B)+2\cos \left(\frac{\pi }{4}-C\right)\cos (A+B)=0\end{array}$$

$$\Rightarrow \sin \left(\frac{\pi }{4}-C+A-B\right)+\sin \left(\frac{\pi }{4}-C-A+B\right)+\cos \left(\frac{\pi }{4}-C+A+B\right)+\cos \left(\frac{\pi }{4}-C-A-B\right)=0\dots \dots$$