Question

Prove the identity $\left[\text{cosec}\right({90}^{\circ }-\theta )-\sin ({90}^{\circ }-\theta \left)\right][\text{cosec}\theta -\sin \theta ][\tan \theta +\cot \theta ]=1$

Answer 1

Now, $\left[\text{cosec}\right({90}^{\circ }-\theta )-\sin ({90}^{\circ }-\theta \left)\right][\text{cosec}\theta -\sin \theta ][\tan \theta +\cot \theta ]$

$=(\sec \theta -\cos \theta )(\text{cosec}\theta -\sin \theta )$
$(\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta })$

$\because \text{cosec}({90}^{\circ }-\theta )=sec\theta$

$\because \sin ({90}^{\circ }-\theta )=\cos \theta$

$=(\frac{1}{\cos \theta }-\cos \theta )(\frac{1}{\sin \theta }-\sin \theta )\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }\right)$

$=\left(\frac{1-{\cos }^2\theta }{\cos \theta }\right)\left(\frac{1-{\sin }^2\theta }{\sin \theta }\right)\left(\frac{1}{\sin \theta \cos \theta }\right)$

$=\left(\frac{{\sin }^2\theta }{\cos \theta }\right)\left(\frac{{\cos }^2\theta }{\sin \theta }\right)\left(\frac{1}{\sin \theta \cos \theta }\right)=1$