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,

& \Rightarrow \tan {{160}^{\circ }}=-\tan {{20}^{\circ }} \\\ & \Rightarrow \tan {{110}^{\circ }}=-\cot {{20}^{\circ }} \\\ \end{aligned}$$ Let us replace $$\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 }}$$.