Search any question & find its solution
Question:
Answered & Verified by Expert
In $\triangle \mathrm{ABC}$, if $\cos \mathrm{A} \cdot \cos \mathrm{B} \cdot \cos \mathrm{C}=\frac{1}{5}$, then $\tan \mathrm{A} \tan \mathrm{B}+$ $\tan \mathrm{B} \tan \mathrm{C}+\tan \mathrm{C} \tan \mathrm{A}=$
Options:
Solution:
1484 Upvotes
Verified Answer
The correct answer is:
6
Given :- $\cos A \cdot \cos B \cdot \cos C=\frac{1}{5}$
$\because \quad A+B+C=\pi \Rightarrow A+B=\pi-C$
$\begin{aligned} & \Rightarrow \quad \cos (A+B)=\cos (\pi-C) \\ & \Rightarrow \quad \cos A \cos B-\sin A \sin B=-\cos C \\ & \Rightarrow \quad \cos A \cos B+\cos C=\sin A \sin B\end{aligned}$
$\Rightarrow \quad 1+\frac{\cos C}{\cos A \cos B}=\tan A \tan B$
or $\tan A \tan B=1+\frac{\cos C}{\cos A \cos B}$ ...(i)
Similarly, we have,
$\tan B \tan C=1+\frac{\cos A}{\cos B \cos C}$ ...(ii)
$\tan C \tan A=1+\frac{\cos B}{\cos C \cos A}$ ...(iii)
Adding eqn. (i), (ii) and (iii), we get
$\tan A \tan B+\tan B \tan C+\tan C \tan A$
$=3+\frac{\cos C}{\cos A \cos B}+\frac{\cos A}{\cos B \cos C}+\frac{\cos B}{\cos C \cos A}$
$=3+\frac{\cos ^2 C+\cos ^2 A+\cos ^2 B}{\cos A \cos B \cos C}$
$=3+5\left(\cos ^2 C+\cos ^2 A+\cos ^2 B\right)$ ...(iv)
On $\triangle A B C$,
$\cos ^2 A+\cos ^2 B+\cos ^2 C=1-2 \cos A \cos B \cos C$ ...(v)
Using eqn. (v) in eqn. (iv), we get
$\tan A \tan B+\tan B \tan C+\tan C \tan A$
$=3+5(1-2 \cos A \cos B \cos C)$
$=3+5\left(1-2 \times \frac{1}{5}\right)=3+5 \times \frac{3}{5}=3+3$
$\Rightarrow \quad \tan A \tan B+\tan B \tan C+\tan C \tan A=6$
Option (3) is correct.
$\because \quad A+B+C=\pi \Rightarrow A+B=\pi-C$
$\begin{aligned} & \Rightarrow \quad \cos (A+B)=\cos (\pi-C) \\ & \Rightarrow \quad \cos A \cos B-\sin A \sin B=-\cos C \\ & \Rightarrow \quad \cos A \cos B+\cos C=\sin A \sin B\end{aligned}$
$\Rightarrow \quad 1+\frac{\cos C}{\cos A \cos B}=\tan A \tan B$
or $\tan A \tan B=1+\frac{\cos C}{\cos A \cos B}$ ...(i)
Similarly, we have,
$\tan B \tan C=1+\frac{\cos A}{\cos B \cos C}$ ...(ii)
$\tan C \tan A=1+\frac{\cos B}{\cos C \cos A}$ ...(iii)
Adding eqn. (i), (ii) and (iii), we get
$\tan A \tan B+\tan B \tan C+\tan C \tan A$
$=3+\frac{\cos C}{\cos A \cos B}+\frac{\cos A}{\cos B \cos C}+\frac{\cos B}{\cos C \cos A}$
$=3+\frac{\cos ^2 C+\cos ^2 A+\cos ^2 B}{\cos A \cos B \cos C}$
$=3+5\left(\cos ^2 C+\cos ^2 A+\cos ^2 B\right)$ ...(iv)
On $\triangle A B C$,
$\cos ^2 A+\cos ^2 B+\cos ^2 C=1-2 \cos A \cos B \cos C$ ...(v)
Using eqn. (v) in eqn. (iv), we get
$\tan A \tan B+\tan B \tan C+\tan C \tan A$
$=3+5(1-2 \cos A \cos B \cos C)$
$=3+5\left(1-2 \times \frac{1}{5}\right)=3+5 \times \frac{3}{5}=3+3$
$\Rightarrow \quad \tan A \tan B+\tan B \tan C+\tan C \tan A=6$
Option (3) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.