Question

Find the value of the expression
$\tan 6{}^\circ \tan 42{}^\circ \tan 66{}^\circ \tan 78{}^\circ$
[a] 0
[b] $\dfrac{1}{2}$
[c] 1
[d] -1

Answer 1

Hint: Observe that $66{}^\circ =180{}^\circ -114{}^\circ$ and $6{}^\circ =60{}^\circ -54{}^\circ$ $114{}^\circ =60{}^\circ +54{}^\circ$. Also, observe that $42{}^\circ =60{}^\circ -18{}^\circ$ and $78{}^\circ =60{}^\circ +18{}^\circ$. Hence prove that the given product is equal to $-\tan \left( 60{}^\circ -54{}^\circ \right)\tan \left( 60{}^\circ +54{}^\circ \right)\tan \left( 60{}^\circ -18{}^\circ \right)\tan \left( 60{}^\circ +18{}^\circ \right)$. Use the fact that $\tan \left( 60{}^\circ -A \right)\tan A\tan \left( 60{}^\circ +A \right)=\tan 3A$ and hence find the value of the given expression.

Complete step-by-step answer:
Let $P=\tan 6{}^\circ \tan 42{}^\circ \tan 66{}^\circ \tan 78{}^\circ$
We know that $\tan \left( x \right)=-\tan \left( 180{}^\circ -x \right)$
Hence, we have
$P=-\tan 6{}^\circ \tan 42{}^\circ \tan \left( 180{}^\circ -66{}^\circ \right)\tan 78{}^\circ$
Hence, we have
$P=-\tan 6{}^\circ \tan 42{}^\circ \tan \left( 114{}^\circ \right)\tan 78{}^\circ$
We know that $66{}^\circ =180{}^\circ -114{}^\circ$ and $6{}^\circ =60{}^\circ -54{}^\circ$ $114{}^\circ =60{}^\circ +54{}^\circ$. Also, observe that $42{}^\circ =60{}^\circ -18{}^\circ$ and $78{}^\circ =60{}^\circ +18{}^\circ$.
Hence, we have
$P=-\tan \left( 60{}^\circ -54{}^\circ \right)\tan \left( 60{}^\circ -18{}^\circ \right)\tan \left( 60{}^\circ +54{}^\circ \right)\tan \left( 60{}^\circ +18{}^\circ \right)$
Rewriting, we get
$P=-\tan \left( 60{}^\circ -54{}^\circ \right)\tan \left( 60{}^\circ +54{}^\circ \right)\tan \left( 60{}^\circ -18{}^\circ \right)\tan \left( 60{}^\circ +18{}^\circ \right)$
Now, we know that
$\tan \left( 60{}^\circ -A \right)\tan A\tan \left( 60{}^\circ +A \right)=\tan 3A$
Dividing both sides of the equation by tanA, we get
$\tan \left( 60{}^\circ -A \right)\tan \left( 60{}^\circ +A \right)=\dfrac{\tan 3A}{\tan A}$
Hence, we have
$\tan \left( 60{}^\circ -54{}^\circ \right)\tan \left( 60{}^\circ +54{}^\circ \right)=\dfrac{\tan \left( 3\times 54{}^\circ \right)}{\tan 54{}^\circ }=\dfrac{\tan 162{}^\circ }{\tan 54{}^\circ }$
We know that
$\tan \left( x \right)=-\tan \left( 180{}^\circ -x \right)$
Hence, we have
$\tan \left( 60{}^\circ -54{}^\circ \right)\tan \left( 60{}^\circ +54{}^\circ \right)=-\dfrac{\tan \left( 180{}^\circ -162{}^\circ \right)}{\tan 54{}^\circ }=-\dfrac{\tan 18{}^\circ }{\tan 54{}^\circ }$
Also, we have
$\tan \left( 60{}^\circ -18{}^\circ \right)\tan \left( 60{}^\circ +18{}^\circ \right)=\dfrac{\tan \left( 3\times 18{}^\circ \right)}{\tan 18{}^\circ }=\dfrac{\tan 54{}^\circ }{\tan 18{}^\circ }$
Hence, we have
$P=-\tan \left( 60{}^\circ -54{}^\circ \right)\tan \left( 60{}^\circ +54{}^\circ \right)\tan \left( 60{}^\circ -18{}^\circ \right)\tan \left( 60{}^\circ +18{}^\circ \right)=-\left( -\dfrac{\tan 18{}^\circ }{\tan 54{}^\circ } \right)\times \dfrac{\tan 54{}^\circ }{\tan 18{}^\circ }$
Hence, we have
P = 1.
Hence option [c] is correct.

Note: [1] In the above question, we have used the property that $\tan \left( 60{}^\circ -A \right)\tan A\tan \left( 60{}^\circ +A \right)=\dfrac{\tan 3A}{\tan A}$
This can be proved as follows:
We know that
$\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$
Replace A by 60 and B by A, we get
$\tan \left( 60{}^\circ -A \right)=\dfrac{\tan 60{}^\circ -\tan A}{1+\tan 60{}^\circ \tan A}=\dfrac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}\text{ }\left( i \right)$
We know that $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
Replace A by 60 and B by A, we get
$\tan \left( 60{}^\circ +A \right)=\dfrac{\tan 60{}^\circ +\tan A}{1-\tan 60{}^\circ \tan A}=\dfrac{\sqrt{3}+\tan A}{1-\sqrt{3}\tan A}\text{ }\left( ii \right)$
Multiplying equation (i) and equation (ii), we get
$\tan \left( 60{}^\circ -A \right)\tan \left( 60{}^\circ +A \right)=\dfrac{\sqrt{3}+\tan A}{1-\sqrt{3}\tan A}\times \dfrac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Using the above identity, we get
$\tan \left( 60{}^\circ -A \right)\tan \left( 60{}^\circ +A \right)=\dfrac{3-{{\tan }^{2}}A}{1-3\tan A}$
Multiplying and dividing by tanA on RHS, we get
$\tan \left( 60{}^\circ -A \right)\tan \left( 60{}^\circ +A \right)=\dfrac{1}{\tan A}\dfrac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}$
We know that $\tan 3A=\dfrac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}$
Hence, we have
$\tan \left( 60{}^\circ -A \right)\tan \left( 60{}^\circ +A \right)=\dfrac{\tan 3A}{\tan A}$, which proves the result
[2] Rule for converting $T\left( n\dfrac{\pi }{2}\pm x \right)$ to $T\left( x \right)$, where T is any trigonometric ratio.
If n is even the final expression will be of T. If n is odd, the final expression will contain the complement of T.
The complement of sin is cos and vice versa
The complement of tan is cot and vice versa
The complement of cosec is sec and vice versa.
Sign of the final expression is determined by the quadrant in which $n\dfrac{\pi }{2}\pm x$ falls.
Keeping the above points in consideration, we have
$\tan \left( \pi -x \right)=\tan \left( 2\dfrac{\pi }{2}-x \right)$
Now 2 is even, hence the final expression will be of tan x.
Also, $\pi -x$ falls in the second quadrant in which tanx is negative
Hence, we have
$\tan \left( \pi -x \right)=-\tan x$