Question

If $\tan 25{}^\circ =x$ , prove that $\dfrac{\tan 155{}^\circ -\tan 115{}^\circ }{1+\tan 155{}^\circ \tan 115{}^\circ }=\dfrac{1-{{x}^{2}}}{2x}$ .

Answer 1

To prove $\dfrac{\tan 155{}^\circ -\tan 115{}^\circ }{1+\tan 155{}^\circ \tan 115{}^\circ }=\dfrac{1-{{x}^{2}}}{2x}$ , we have to consider the LHS. We have to change the angles in such a way that $25{}^\circ$ will be included, that is, we have to write $155{}^\circ =\left( 180{}^\circ -25{}^\circ \right)$ and $115{}^\circ =\left( 90{}^\circ +25{}^\circ \right)$ . We have to substitute these in the LHS and apply the trigonometric rules $\tan \left( 180{}^\circ -\theta \right)=-\tan \theta$ and $\tan \left( 90{}^\circ +\theta \right)=-\cot \theta$ . Then, we have to use the given condition and simplify.

Complete step by step solution:
We have to prove $\dfrac{\tan 155{}^\circ -\tan 115{}^\circ }{1+\tan 155{}^\circ \tan 115{}^\circ }=\dfrac{1-{{x}^{2}}}{2x}$ . Let us consider the LHS.
$\Rightarrow LHS=\dfrac{\tan 155{}^\circ -\tan 115{}^\circ }{1+\tan 155{}^\circ \tan 115{}^\circ }$
We know that $155{}^\circ =\left( 180{}^\circ -25{}^\circ \right)$ and $115{}^\circ =\left( 90{}^\circ +25{}^\circ \right)$ . Hence, we can write the LHS as
$\Rightarrow \dfrac{\tan 155{}^\circ -\tan 115{}^\circ }{1+\tan 155{}^\circ \tan 115{}^\circ }=\dfrac{\tan \left( 180{}^\circ -25{}^\circ \right)-\tan \left( 90{}^\circ +25{}^\circ \right)}{1+\tan \left( 180{}^\circ -25{}^\circ \right)\tan \left( 90{}^\circ +25{}^\circ \right)}$
We know that $\tan \left( 180{}^\circ -\theta \right)=-\tan \theta$ and $\tan \left( 90{}^\circ +\theta \right)=-\cot \theta$ . Therefore, the above equation becomes

& \Rightarrow \dfrac{\tan 155{}^\circ -\tan 115{}^\circ }{1+\tan 155{}^\circ \tan 115{}^\circ }=\dfrac{-\tan 25{}^\circ -\left( -\cot 25{}^\circ \right)}{1-\tan 25{}^\circ \times -\cot 25{}^\circ } \\\ & \Rightarrow \dfrac{\tan 155{}^\circ -\tan 115{}^\circ }{1+\tan 155{}^\circ \tan 115{}^\circ }=\dfrac{-\tan 25{}^\circ +\cot 25{}^\circ }{1+\tan 25{}^\circ \cot 25{}^\circ }...\left( i \right) \\\ \end{aligned}$$ We are given that $\tan 25{}^\circ =x...\left( ii \right)$ . We know that $\cot \theta =\dfrac{1}{\tan \theta }$ . Therefore, we can find $\cot 25{}^\circ $ as $\cot 25{}^\circ =\dfrac{1}{\tan 25{}^\circ }=\dfrac{1}{x}...\left( iii \right)$ Let us substitute (ii) and (iii) in (i). $$\Rightarrow \dfrac{\tan 155{}^\circ -\tan 115{}^\circ }{1+\tan 155{}^\circ \tan 115{}^\circ }=\dfrac{-x+\dfrac{1}{x}}{1+x\times \dfrac{1}{x}}$$ Let us take the LCM and solve. $$\begin{aligned} & \Rightarrow \dfrac{\tan 155{}^\circ -\tan 115{}^\circ }{1+\tan 155{}^\circ \tan 115{}^\circ }=\dfrac{\dfrac{-{{x}^{2}}+1}{x}}{1+1} \\\ & \Rightarrow \dfrac{\tan 155{}^\circ -\tan 115{}^\circ }{1+\tan 155{}^\circ \tan 115{}^\circ }=\dfrac{1-{{x}^{2}}}{2x}=RHS \\\ \end{aligned}$$ We can see that LHS = RHS. Hence proved. **Note:** Students must know the trigonometric formulas and rules to solve these problems. They must know that all the trigonometric functions are positive in the angle $\left( 90{}^\circ -\theta \right)$ and the function changes. If the function was sin, it would be cos and vise-versa, tan to cot and vise-versa, cosec to sec and vise-versa. Only sine and cosec will be positive when the angle is $\left( 90{}^\circ +\theta \right)$ or $\left( 180{}^\circ -\theta \right)$ . When the angle is $\left( 90{}^\circ +\theta \right)$ , the sine function changes to cosine and cosine function to sine (with a negative sign as cosine is negative in this angle). The cosec function changes to sec at this angle and sec changes to cosec by with a negative sign. Only tangent and cotangent functions will be positive when angle is $\left( 180{}^\circ +\theta \right)$ or $\left( 270{}^\circ -\theta \right)$ . The function tan changes to cot and vise-versa when the angle is $\left( 270{}^\circ -\theta \right)$ . When the angle is $\left( 270{}^\circ +\theta \right)$ or $\left( 360{}^\circ -\theta \right)$ , cos and sec functions are positive. The function cos changes to sin at the angle $\left( 270{}^\circ +\theta \right)$ and also sec changes to cosec. The given figure describes this.