If α , β are different values of x satisfying acos x + bsin x = c, then tan {rm{ }}left( {frac{{α

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Trigonometrical Equations

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If $\alpha ,$ $\beta$ are different values of $x$ satisfying $a\cos x + b\sin x = c,$ then $\tan {\rm{ }}\left( {\frac{{\alpha + \beta }}{2}} \right) = $

$a + b$

$a - b$

$\frac{b}{a}$

$\frac{a}{b}$

Solution

$ \Rightarrow $ $a\,\left( {\frac{{1 – {{\tan }^2}\,(x/2)}}{{1 + {{\tan }^2}(x/2)}}} \right) + \frac{{2b\tan (x/2)}}{{1 + {{\tan }^2}(x/2)}} = c$

$ \Rightarrow $ $(a + c){\tan ^2}\frac{x}{2} – 2b\tan \frac{x}{2} + (c – a) = 0$

This equation has roots $\tan \frac{\alpha }{2}$ and $\tan \frac{\beta }{2}$.

$\therefore $ $\tan \frac{\alpha }{2} + \tan \frac{\beta }{2} = \frac{{2b}}{{a + c}}$ and

$\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{{c – a}}{{a + c}}$

Now $\tan \left( {\frac{\alpha }{2} + \frac{\beta }{2}} \right) = \frac{{\tan \frac{\alpha }{2} + \tan \frac{\beta }{2}}}{{1 – \tan \frac{\alpha }{2}\tan \frac{\beta }{2}}}$

$= \frac{{\frac{{2b}}{{a + c}}}}{{1 – \frac{{c – a}}{{a + c}}}} = \frac{b}{a}$.

Standard 11

Mathematics

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If $\alpha ,$ $\beta$ are different values of $x$ satisfying $a\cos x + b\sin x = c,$ then $\tan {\rm{ }}\left( {\frac{{\alpha + \beta }}{2}} \right) = $

Solution

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