Question

If $\theta$ and $\sin \theta =\frac{a}{\sqrt{a^2+b^2}}$, the value of the expression $a\text{cosec}\theta -b\sec \theta$ is

Answer 1

$sin\theta =\left)\frac{a}{\sqrt{a^2+b^2}}\right)\therefore cos\theta \sqrt{1-si{n}^2\theta }=\sqrt{1-\left)\frac{a^2}{a^2+b^2}\right)}=\sqrt{\left)\frac{a^2+b^2-a^2}{a^2+b^2}\right)}=\left)\frac{b}{\sqrt{a^2+b^2}}\right)\therefore acosec\theta -bsec\theta =\left)\frac{a}{\left)\frac{a}{\sqrt{a^2+b^2}}\right)}\right)-\left)\frac{b}{\left)\frac{b}{\sqrt{a^2+b^2}}\right)}\right)=\sqrt{a^2+b^2}-\sqrt{a^2+b^2}=0$