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  \(\begin{align}\frac{B}{2} \leqslant \frac{\pi }{6}  \Rightarrow B \leqslant \frac{\pi }{3}\end{align}\)

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