Trigonometry Equations – 1

\displaystyle Q.1) Prove:{{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \frac{a+b}{1-ab} \right)

 

Solution :

   \displaystyle \text{Let }{{\tan }^{-1}}a=\alpha ,\text{ so that tan}\alpha \text{  =  a}

\displaystyle {{\tan }^{-1}}b=\beta ,\text{ so that tan}\beta \text{ = b}

\displaystyle \text{We have}\tan \left( \alpha +\beta  \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }

 

\displaystyle =\frac{a+b}{1-ab}

 

\displaystyle \text{ Now, }\left( \alpha +\beta  \right)={{\tan }^{-1}}\left( \frac{a+b}{1-ab} \right)

\displaystyle \text{After Re-substitution, we get}

\displaystyle {{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \frac{a+b}{1-ab} \right)

\displaystyle \text{Hence proved}\text{.}

 

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