If \(A B C\) is not a right angled triangle and \(\sin \left(\frac{\pi}{4}-A\right) \sin…
If \(A B C\) is not a right angled triangle and \(\sin \left(\frac{\pi}{4}-A\right) \sin \left(\frac{\pi}{4}-B\right)=-\frac{1}{2 \sqrt{2}} \operatorname{cosec}\left(\frac{\pi}{4}-C\right)\), then \(\tan A \tan B+\tan B \tan C+\tan C \tan A=\)
\(\cot A+\cot B+\cot C\)
\(\tan A+\tan B+\tan C\)
\(\frac{1}{\tan A+\tan B+\tan C}\)
\(\frac{1}{\cot A+\cot B+\cot C}\)
Solution
Given, \(A+B+C=\pi\)
We know that,
\(\cos (A+B+C)=\cos A \cos B \cos C(l-\tan A \tan B\)
\(-\tan B \tan C-\tan A \tan C)\)
and \(\tan A+\tan B+\tan C=\tan A \tan B \tan C\)
Now, \(\sin \left(\frac{\pi}{4}-A\right) \sin \left(\frac{\pi}{4}-B\right) \sin \left(\frac{\pi}{4}-C\right)=-\frac{1}{2 \sqrt{2}}\)
\(\Rightarrow\left(\frac{1}{\sqrt{3}}\right)^3(\cos A-\sin A)(\cos B-\sin B)\)
\((\cos C-\sin C)=-\frac{1}{2 \sqrt{2}}\)
\(\Rightarrow \cos A \cos B \cos C(1-\tan A)(1-\tan B)\)
\((1-\tan C)=-1\)
\(\Rightarrow \cos A \cos B \cos C(1-\tan A-\tan B-\tan C\) \(+\tan A \tan B+\tan B \tan C+\tan A \tan C\)
\(-\tan A \tan B \tan C)\)
\(=\cos A \cos B \cos C(1-\tan A \tan B\)
\(-\tan B \tan C-\tan A \tan C)\)
\(=\cos (A+B+C)\)
\((\because A+B+C=\pi)\)
\(\Rightarrow 1-2(\tan A+\tan B+\tan C)+\tan A \tan B\)
\(+\tan B \tan C+\tan A \tan C\)
\(=\mathrm{l}-\tan A \tan B-\tan B \tan C-\tan A \tan C\)
\(\Rightarrow-2(\tan A+\tan B+\tan C)=-2(\tan A \tan B\)
\(+\tan B \tan C+\tan A \tan C)\)
\(\Rightarrow \tan A \tan B+\tan B \tan C+\tan A \tan C\)
\(=\tan A+\tan B+\tan C\)