Question

If \[\tan {{20}^{\circ }}=\lambda \], then show that \[\dfrac{\tan {{160}^{\circ }}-\tan {{110}^{\circ }}}{1+\tan {{160}^{\circ }}.\tan {{110}^{\circ }}}=\dfrac{1-{{\lambda }^{2}}}{2\lambda }\].

Answer 1

Hint: Find the value of \[\tan {{160}^{\circ }}\] & \[\tan {{110}^{\circ }}\] in terms of \[\tan {{20}^{\circ }}\] using the trigonometric functions \[\tan \left( 180-\theta \right)\] and \[\tan \left( 90+\theta \right)\]. Replace the values in L.H.S of the expression, put \[\tan {{20}^{\circ }}=\lambda \] and simplify it.

Complete step-by-step answer:
We are given the tangent function, \[\tan {{20}^{\circ }}=\lambda \].
We need to show that,
\[\Rightarrow \dfrac{\tan {{160}^{\circ }}-\tan {{110}^{\circ }}}{1+\tan {{160}^{\circ }}.\tan {{110}^{\circ }}}=\dfrac{1-{{\lambda }^{2}}}{2\lambda }\]
We know the basic trigonometric function such as,
\[\Rightarrow \tan \left( 180-\theta \right)=-\tan \theta \]
Similarly, \[\tan \left( 90+\theta \right)=-\cot \theta \].
Hence we can change the expression, \[\tan {{160}^{\circ }}\] and \[\tan {{110}^{\circ }}\].
\[\tan {{160}^{\circ }}\] can be written as, \[\tan \left( 180-{{20}^{\circ }} \right)\].
i.e. \[\tan {{160}^{\circ }}\] = \[\tan \left( {{180}^{\circ }}-{{20}^{\circ }} \right)=-\tan {{20}^{\circ }}\].
Similarly \[\tan {{110}^{\circ }}\] can be written as \[\tan \left( {{90}^{\circ }}+{{20}^{\circ }} \right)\].
i.e. \[\tan {{110}^{\circ }}\] = \[\tan \left( {{90}^{\circ }}+{{20}^{\circ }} \right)\] = \[-\cot {{20}^{\circ }}\]
There we can say that,
\[\begin{align}
  & \Rightarrow \tan {{160}^{\circ }}=-\tan {{20}^{\circ }} \\
 & \Rightarrow \tan {{110}^{\circ }}=-\cot {{20}^{\circ }} \\
\end{align}\]
Let us replace \[\tan {{160}^{\circ }}\] and \[\tan {{110}^{\circ }}\] in the LHS of the expression,
LHS = \[\dfrac{\tan {{160}^{\circ }}-\tan {{110}^{\circ }}}{1+\left( \tan {{160}^{\circ }} \right).\left( \tan {{110}^{\circ }} \right)}=\dfrac{-\tan {{20}^{\circ }}-\left( -\cot {{20}^{\circ }} \right)}{1+\left( -\tan {{20}^{\circ }} \right)\left( -\cot {{20}^{\circ }} \right)}\]
We know that, \[\cot {{20}^{\circ }}=\dfrac{1}{\tan {{20}^{\circ }}}\].
\[\therefore \] LHS = \[\dfrac{-\tan {{20}^{\circ }}+\cot {{20}^{\circ }}}{1+\tan {{20}^{\circ }}\times \dfrac{1}{\tan {{20}^{\circ }}}}=\dfrac{-\tan {{20}^{\circ }}+\dfrac{1}{\tan {{20}^{\circ }}}}{1+1}\]
Let us put \[\tan {{20}^{\circ }}=\lambda \], in the above expression.
L.H.S = \[\dfrac{-\lambda +\dfrac{1}{\lambda }}{2}=\dfrac{-{{\lambda }^{2}}+1}{2\lambda }=\dfrac{1-{{\lambda }^{2}}}{2\lambda }\]
Thus we got LHS = \[\dfrac{1-{{\lambda }^{2}}}{2\lambda }\], which is equal to the RHS.
Thus we can say that, LHS = RHS.
i.e. We prove that they are equal.
\[\therefore \dfrac{\tan {{160}^{\circ }}-\tan {{110}^{\circ }}}{1+\tan {{160}^{\circ }}.\tan {{110}^{\circ }}}=\dfrac{1-{{\lambda }^{2}}}{2\lambda }\]
Hence proved

Note: By seeing the LHS of the expression, you may assume that it is the expansion of the formula, \[\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}\]. But \[\tan \left( {{160}^{\circ }}-{{110}^{\circ }} \right)=\tan {{50}^{\circ }}\], it won’t give us the desirable results. Then use trigonometric identity \[\tan \left( 180-\theta \right)\] and \[\tan \left( 90+\theta \right)\] and convert the terms of \[\tan {{20}^{\circ }}\].