Question

If $\frac{\cos ({90}^0+\theta ).\sec ({270}^0+\theta ).\sin ({180}^0+\theta )}{\cos (-\theta ).\cos ({270}^0-\theta ).\tan ({180}^0+\theta )}=-m\text{cosec}\theta$.Find $m$

Answer 1

LHS- $\frac{\cos (90+\theta ).\sec (270+\theta ).\sin (180+\theta )}{\cos (-\theta ).\cos (270-\theta ).\tan (180+\theta )}$

$\Rightarrow \frac{(-\sin \theta )(+\csc \theta )(-\sin \theta )}{\left(\cos \theta \right)(-\sin \theta )\left(\tan \theta \right)}$

$\Rightarrow \frac{-\left(\sin \theta \right)\left(\frac{1}{\sin \theta }\right)}{\left(\cos \theta \right)\left(\frac{\sin \theta }{\cos \theta }\right)}$

$\Rightarrow \frac{-1}{\sin \theta }$

$\Rightarrow -\csc \theta$

$\Rightarrow -m\csc \theta$

$\Rightarrow m=1$.