Question 12, Exercise 8.1

Solutions of Question 12 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

If $\alpha+\beta+\gamma=180^{\circ}$, prove that: $\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma$.

Solution.

Given: $$\alpha+\beta+\gamma=180^{\circ}$$ This gives $$\begin{align*} & \alpha+\beta=180^{\circ}-\gamma \\ \implies & \tan(\alpha+\beta) = \tan(180^{\circ}-\gamma) \\ \implies & \frac{\tan\alpha + \tan\beta}{1-\tan\alpha \tan\beta} = \tan(2(90)-\gamma) \\ \implies & \tan\alpha + \tan\beta = -\tan\gamma[1-\tan\alpha \tan\beta] \\ \implies & \tan\alpha + \tan\beta = -\tan\gamma+\tan\alpha \tan\beta\tan\gamma \\ \implies & \tan\alpha + \tan\beta +\tan\gamma=\tan\alpha \tan\beta\tan\gamma \\ \end{align*}$$ as required.

If $\alpha+\beta+\gamma=180^{\circ}$, prove that: $\cot \frac{\alpha}{2}+\cot \frac{\beta}{2}+\cot \frac{\gamma}{2}=\cot \frac{\alpha}{2} \cot \frac{\beta}{2} \cot \frac{\gamma}{2}$

Solution.

Given: $$\alpha+\beta+\gamma=180^{\circ}$$ This gives $$\begin{align*} & \alpha+\beta=180^{\circ}-\gamma \\ \implies & \frac{\alpha +\beta }{2}=\frac{180-\gamma }{2} \\ \implies & \frac{\alpha }{2}+\frac{\beta }{2}=90-\frac{\gamma }{2} \end{align*}$$ Now $$\begin{align*} & \tan \left( \frac{\alpha }{2}+\frac{\beta }{2} \right)=\tan \left( 90-\frac{\gamma }{2} \right) \\ \implies & \frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}=\cot \frac{\gamma }{2} \qquad \because \tan \left( 90-\frac{\gamma }{2} \right)=\cot \frac{\gamma }{2} \\ \implies & \frac{\tan \frac{\alpha }{2}\,\tan \frac{\beta }{2}\left( \frac{1}{\tan \tfrac{\beta }{2}}+\frac{1}{\tan \tfrac{\alpha }{2}} \right)}{\tan \frac{\alpha }{2}\,\tan \frac{\beta }{2}\left( \frac{1}{\tan \tfrac{\alpha }{2}\tan \tfrac{\beta }{2}}-1 \right)}=\cot \frac{\gamma }{2} \\ \implies & \frac{\cot \frac{\beta }{2}+\cot \frac{\alpha }{2}}{\cot \frac{\alpha }{2}\cot \frac{\beta }{2}-1}=\cot \frac{\gamma }{2} \\ \implies & \cot \frac{\beta }{2}+\cot \frac{\alpha }{2}=\cot \frac{\gamma }{2}\left( \cot \frac{\alpha }{2}\cot \frac{\beta }{2}-1 \right) \\ \implies & \cot \frac{\beta }{2}+\cot \frac{\alpha }{2}=\cot \frac{\alpha }{2}\cot \frac{\beta }{2}\cot \frac{\gamma }{2}\,\,-\,\,\cot \frac{\gamma }{2}\\ \implies & \cot \frac{\beta }{2}+\cot \frac{\alpha }{2}+\,\,\cot \frac{\gamma }{2}=\cot \frac{\alpha }{2}\cot \frac{\beta }{2}\cot \frac{\gamma }{2} \end{align*}$$ as required.

If $\alpha+\beta+\gamma=180^{\circ}$, prove that: $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}+\tan \frac{\beta}{2} \tan \frac{\gamma}{2}+\tan \frac{\gamma}{2} \tan \frac{\alpha}{2}+1=0$

Solution.

Given: $$\alpha+\beta+\gamma=180^{\circ}$$ This gives $$\begin{align*} & \alpha+\beta=180^{\circ}-\gamma \\ \implies & \frac{\alpha +\beta }{2}=\frac{180-\gamma }{2} \\ \implies & \frac{\alpha }{2}+\frac{\beta }{2}=90-\frac{\gamma }{2} \end{align*}$$ Now $$\begin{align*} & \tan \left( \frac{\alpha }{2}+\frac{\beta }{2} \right)=\tan \left( 90-\frac{\gamma }{2} \right) \\ \implies & \frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}=\cot \frac{\gamma }{2} \\ \implies & \frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{\cot \frac{\gamma }{2}}= 1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2} \\ \implies & \left(\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}\right) \tan \frac{\gamma }{2}= 1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2} \\ \implies & \tan \frac{\gamma }{2} \tan \frac{\alpha }{2}+\tan \frac{\beta }{2}\tan \frac{\gamma }{2}= 1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2} \\ \implies & \tan \frac{\alpha }{2}\tan \frac{\beta }{2}+\tan \frac{\beta }{2}\tan \frac{\gamma }{2}+\tan \frac{\gamma }{2} \tan \frac{\alpha }{2}-1= 0 \\ \end{align*}$$

as required.

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