# Question11 and 12, Exercise 10.1

Solutions of Question 11 and 12 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If $\alpha$, $\beta$, $\gamma$ are the angles of a triangle $ABC$, show that $\cot \dfrac{\alpha }{2}+\cot \dfrac{\beta }{2}+\cot \dfrac{\gamma }{2}=\cot \dfrac{\alpha }{2}\cot \dfrac{\beta }{2}\cot \dfrac{\gamma }{2}$

Since $\alpha$, $\beta$ and $\gamma$ are angles of triangle, therefore \begin{align}&\alpha+\beta +\gamma =180^\circ\\ \implies &\alpha +\beta =180^\circ-\gamma \\ \implies &\dfrac{\alpha +\beta }{2}=\dfrac{180^\circ-\gamma }{2}\\ \implies &\dfrac{\alpha}{2}+\dfrac{\beta}{2}=90^\circ-\dfrac{\gamma}{2}\end{align} Now \begin{align}&\tan\left( \dfrac{\alpha }{2}+\dfrac{\beta }{2}\right)=\tan \left( 90-\dfrac{\gamma }{2} \right)\\ \implies &\dfrac{\tan\dfrac{\alpha}{2}+\tan\dfrac{\beta}{2}}{1-\tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2}}=\cot \dfrac{\gamma}{2} \quad \because\,\,\tan\left( 90-\tfrac{\gamma}{2}\right)=\cot\tfrac{\gamma}{2}\end{align} \begin{align} \implies & \frac{\tan\dfrac{\alpha}{2}\tan\dfrac{\beta}{2}\left(\dfrac{1}{\tan\tfrac{\beta}{2}}+\dfrac{1}{\tan \tfrac{\alpha}{2}}\right)}{\tan\dfrac{\alpha}{2}\tan\dfrac{\beta}{2}\left(\dfrac{1}{\tan\tfrac{\alpha}{2}\tan \tfrac{\beta}{2}}-1\right)}=\cot\dfrac{\gamma}{2}\\ \implies &\dfrac{\cot\dfrac{\beta}{2}+\cot\dfrac{\alpha}{2}}{\cot\dfrac{\alpha}{2}\cot\dfrac{\beta }{2}-1}=\cot \dfrac{\gamma }{2}\\ \implies &\cot\dfrac{\beta}{2}+\cot\dfrac{\alpha}{2}=\cot\dfrac{\gamma}{2}\left(\cot\dfrac{\alpha }{2}\cot \dfrac{\beta}{2}-1 \right)\\ \implies &\cot\dfrac{\beta}{2}+\cot\dfrac{\alpha}{2}=\cot\dfrac{\alpha}{2}\cot\dfrac{\beta}{2}\cot\dfrac{\gamma}{2}-\cot\dfrac{\gamma}{2}\\ \implies &\cot\dfrac{\alpha}{2}+\cot\dfrac{\beta}{2}+\cot\dfrac{\gamma}{2}=\cot\dfrac{\alpha}{2}\cot\dfrac{\beta}{2}\cot\dfrac{\gamma}{2}.\end{align}

If $\alpha +\beta +\gamma ={{180}^{\circ}}$, show that $\cot\alpha \cot \beta +\cot \beta \cot \gamma +\cot \gamma \cot \alpha =1$

Since $\alpha,\quad \beta$ and $\gamma$ are angles of triangle, therefore \begin{align}&\alpha +\beta +\gamma =180^\circ\\ \implies &\alpha +\beta=180^\circ-\gamma. \end{align} Now \begin{align} & \tan \left( \alpha +\beta \right)=\tan \left( 180^\circ-\gamma \right)\\ \implies &\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\tan \left( 2(90^\circ)-\gamma \right)\\ \implies &\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=-\tan \gamma \end{align}\begin{align} \implies &\tan \alpha +\tan \beta =-\tan \gamma \left( 1-\tan \alpha \tan \beta \right)\\ \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} Dividing through out by $\tan \alpha \tan \beta \tan \gamma$ to get \begin{align} &\dfrac{\tan \alpha }{\tan \alpha \tan \beta \tan \gamma }+\dfrac{\tan \beta }{\tan \alpha \tan \beta \tan \gamma }+\dfrac{\tan \gamma }{\tan \alpha \tan \beta \tan \gamma }=1\\ \implies &\cot \beta \cot \gamma +\cot \alpha \cot \gamma +\cot \alpha \cot \beta =1\\ \implies &\cot \alpha \cot \beta +\cot \beta \cot \gamma +\cot \gamma \cot \alpha =1.\end{align}