# Examples on Conditional Trigonometric Identities Set 2

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Example - 29

Prove that  $$\Delta ABC$$  is equilateral if and only if

$\tan A + \tan B + \tan C = 3\sqrt 3$

Solution: This is a two-way implication of which one is very easy to prove. Namely, if $$\Delta ABC$$ is equilateral, then

$\tan A + \tan B + \tan C = \sqrt 3 + \sqrt 3 + \sqrt 3 = 3\sqrt 3$

The main part of this problem is proving the converse.

From the previous example, we have

$\tan A + \tan B + \tan C = \tan A\tan B\tan C$

On the other hand, by the  $${\text{A}}{\text{.M}}\;{\text{–}}\;{\text{G}}{\text{.M}}$$  inequality,

$\tan A + \tan B + \tan C \geqslant {(\tan A\tan B\tan C)^{1/3}}$

The A.M. of $$\tan A,\;\tan B,\;\tan C\;{\text{is}}\;\sqrt 3$$, whereas the G.M $${(3\sqrt 3 )^{1/3}} = \sqrt 3$$. is, i.e., the same as the A.M. This can happen only if

$\tan A = \tan B = \tan C\;\;\;\;\;\;\; \Rightarrow A = B = C$

Example - 30

In a  $$\Delta ABC$$  if  $$\sin 3A + \sin 3B + \sin 3C = 0$$, then show that at least one of the angles is  \begin{align}\frac{\pi }{3}\end{align}.

Solution: Following the approach in previous examples, we can prove that

$\sin 3A + \sin 3B + \sin 3C = - 4\cos \frac{{3A}}{2}\cos \frac{{3B}}{2}\cos \frac{{3C}}{2}$

For this to be zero, one of the cos terms on the RHS must be zero, which means that one of the angles must be  \begin{align}\frac{\pi }{3}\end{align}.

Example - 31

In a   $$\Delta ABC$$, if

$\cos A\cos B\cos C = \frac{{\sqrt 3 - 1}}{8},\;\sin A\sin B\sin C = \frac{{3 + \sqrt 3 }}{8}$

Find the angles of the triangle

Solution: From the given relations,

$\tan A\tan B\tan C = \frac{{3 + \sqrt 3 }}{{\sqrt 3 - 1}}$

Using the result of Example - 28, we have

$\tan A + \tan B + \tan C = \frac{{3 + \sqrt 3 }}{{\sqrt 3 - 1}}$

Now, since \begin{align}A + B + C = \pi \end{align}

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos (A + B + C) = - 1 \\ \Rightarrow\quad &\qquad\;\;\cos A\cos B\cos C\left\{ {1 - \sum (\tan A\tan B)} \right\} = - 1 \\ \Rightarrow\quad &\qquad\;\; \sum \left( {\tan A\tan B} \right) = 1 + \frac{8}{{\sqrt 3 - 1}} = 5 + 4\sqrt 3 \\ \end{align}

Finally, for  $$\tan A,\tan B,\tan C$$, we have their sum, their sum taken pairwise and their product, and thus we can say that these three are the roots of the equation

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^3} - \left( {\frac{{3 + \sqrt 3 }}{{\sqrt 3 - 1}}} \right){x^2} + (5 + 4\sqrt 3 )x - \left( {\frac{{3 + \sqrt 3 }}{{\sqrt 3 - 1}}} \right) = 0\\ \Rightarrow\quad &\qquad\;\;(x - 1)\left( {x - \sqrt 3 } \right)\left( {x - (2 + \sqrt 3 )} \right) = 0 \\ \Rightarrow \quad &\qquad x = 1,\;\sqrt 3 ,\;2 + \sqrt 3 \\ \end{align}

The angles are \begin{align}\frac{\pi }{4},\;\frac{\pi }{3},\;\frac{\pi }{{12}}\end{align}

Example - 32

In a  $$\Delta ABC$$, suppose that  \begin{align}\tan \frac{A}{2},\;\tan \frac{B}{2},\;\tan \frac{C}{2}\end{align}  are in H.P.. What is the minimum possible value of \begin{align}\cot \frac{B}{2}\end{align}?

Solution: Refer to the result of Example - 28 Part (b). If we divide throughout by \begin{align}\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}\end{align},

we have,

$\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)$

Now, since \begin{align}\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2}\end{align}  are in H.P., their reciprocals, i.e. \begin{align}\cot \frac{A}{2},\;\cot \frac{B}{2},\;\cot \frac{C}{2}\end{align}, are in A.P. So,

$\qquad\cot \frac{A}{2} + \;\cot \frac{C}{2} = 2\cot \frac{B}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)$

Using (2) in (1), we obtain

$\cot \frac{A}{2}\;\cot \frac{C}{2} = 3$

Finally, we use the A.M-G.M. inequality on \begin{align}\cot \frac{A}{2},\;\cot \frac{C}{2}\end{align}:

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{\cot \frac{A}{2} + \cot \frac{C}{2}}}{2} = \cot \frac{B}{2} \geqslant \sqrt {\cot \frac{A}{2}\cot \frac{C}{2}} = \sqrt 3 \\ \Rightarrow\quad &\qquad\;\;\;\cot \frac{B}{2} \geqslant \sqrt 3 \\ \end{align}

Therefore, the minimum value of  \begin{align}\cot \frac{B}{2}\end{align}  is  $$\sqrt 3$$. From this, we can further infer that the maximum value of  \begin{align}\tan \frac{B}{2}\end{align} is \begin{align}\frac{1}{{\sqrt 3 }}\end{align}  which implies that  \begin{align}\frac{B}{2} \leqslant \frac{\pi }{6} \Rightarrow B \leqslant \frac{\pi }{3}\end{align}