If and are two solutions of the equation a +b =c, the find the va… - Vidyadip

Question

If $\alpha\text{ and }\beta$ are two solutions of the equation $\text{a}\tan\text{x+b}\sec\text{x}=\text{c},$ the find the value of $\sin(\alpha+\beta)\text{ and }\cos(\alpha+\beta).$

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Answer

$\text{a}\tan\text{x}+\text{b}\sec\text{x}=\text{c}$
$\Rightarrow(\text{c}-\text{a}\tan\text{x})=\text{b}\sec\text{x}$
$\Rightarrow(\text{c}-\text{a}\tan\text{x})^2=(\text{b}\sec\text{x})^2$
$\Rightarrow\text{c}^2+\text{a}^2\tan^2\text{x}-2\text{a}\text{c}\tan\text{x}=\text{b}^2\sec^2\text{x}$
$\Rightarrow\text{c}^2+\text{a}^2\tan^2\text{x}-2\text{a}\text{c}\tan\text{x}=\text{b}^2(1+\tan^2\text{x)}$
$\Rightarrow(\text{a}^2-\text{b}^2)\tan^2\text{x}-2\text{a}\text{c}\tan\text{x}+\text(\text{c}^2-\text{b}^2)=0$
This is a quadratic in $\tan\text{x}.$
It has two solutions $\tan\alpha\text{ and }\tan\beta.$
$\tan\alpha\text{ and }\tan\beta=\frac{2\text{ac}}{\text{a}^2-\text{b}^2}$
$\tan\alpha\times\tan\beta=\frac{\text{c}^2-\text{b}^2}{\text{a}^2-\text{b}^2}$
$\therefore\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$=\frac{\frac{2\text{ac}}{\text{a}^2-\text{b}^2}}{1-\frac{\text{c}^2-\text{b}^2}{\text{a}^2-\text{b}^2}}=\frac{2\text{ac}}{\text{a}^2-\text{c}^2}$
Hence, $\sin(\alpha+\beta)=\frac{ 2\text{ac}}{\text{a}^2-\text{b}^2}\text{ and }\cos(\alpha+\beta)=\frac{\text{a}^2-\text{c}^2}{\text{a}^2+\text{b}^2}.$

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