Question

tanα+ctanα=a find sinα

Answer 1

$$\tan \alpha + c \tan \alpha = a$$ $$\frac {\sin \alpha}{\cos \alpha} + \frac {\cos \alpha}{\sin \alpha} = a$$ $$\frac {\sin^ {2} \alpha + \cos^ {2} \alpha}{\sin \alpha \cos \alpha} = a$$ $$\frac {1}{\sin \alpha \cos \alpha} = a$$ $$\sin \alpha \cos \alpha = \frac {1}{a}$$ $$\sin \alpha \sqrt {1 - \sin^ {2} \alpha} = \frac {1}{a}$$ $$\sin^ {2} \alpha (1 - \sin^ {2} \alpha) = \frac {1}{a ^ {2}}$$ $$\sin^ {4} \alpha - \sin^ {2} \alpha + \frac {1}{a ^ {2}} = 0$$ $$\sin^ {2} \alpha = t$$ $$t ^ {2} - t + \frac {1}{a ^ {2}} = 0$$ $$D = 1 - \frac {4}{a ^ {2}}$$ $$t _ {1} = \frac {- 1 + \sqrt {1 - \frac {4}{a ^ {2}}}}{2} \quad t _ {2} = \frac {- 1 - \sqrt {1 - \frac {4}{a ^ {2}}}}{2} < 0$$ $$\sin \alpha = \pm \sqrt {\frac {- 1 + \sqrt {1 - \frac {4}{a ^ {2}}}}{2}}$$