Question

If $\tan \theta =\frac{a}{b}$ and $0<\theta <\pi$ , then the value of $\therefore \frac{\sqrt{a^2+b^2}}{\mid a\mid }\sin \theta$

Answer 1

$\tan \theta =\frac{a}{b}$ , $0<\theta <\pi$

${\sec }^2\theta -{\tan }^2\theta =1$

$\sec \theta =\sqrt{1+{\tan }^2\theta }$

$\sec \theta =\frac{\sqrt{a^2+b^2}}{\mid b\mid }$

$\cos \theta =\frac{\mid b\mid }{\sqrt{a^2+b^2}}$

$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{b^2}{a^2+b^2}}$

$\sin \theta =\frac{\mid a\mid }{\sqrt{a^2+b^2}}$

$\therefore \frac{\sqrt{a^2+b^2}}{\mid a\mid }\sin \theta =1$