Question

Prove that : $[\csc ({90}^o-\theta )-\sin ({90}^o-\theta )][\csc \theta -\sin \theta ][\tan \theta +\cot \theta ]=1$

Answer 1

$[\csc (90°-\theta )-\sin (90°-\theta )][\csc \theta -\sin \theta ][\tan \theta +\cot \theta ]=1$

$\Rightarrow L.H.S=[\csc (90°-\theta )-\sin (90°-\theta )][\csc \theta -\sin \theta ][\tan \theta +\cot \theta ]$

$=(\sec \theta -\cos \theta )(\csc \theta -\sin \theta )(\tan \theta +\cot \theta )$

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

$=\frac{{\sin }^2\theta }{\cos \theta }.\frac{{\cos }^2\theta }{\sin \theta }(\tan \theta +\cot \theta )$

$=\sin \theta .\cos \theta (\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta })$

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

$={\sin }^2\theta +{\cos }^2\theta$

$=1$

$\Rightarrow L.H.S=R.H.S$

Hence, the answer is proved.