Question

Under root sec^2 theta+csc^2theta=tan theta+cot theta

Answer 1

Dear student
$$\begin{gathered} LHS=\sqrt{se{c}^2\theta +\cos e{c}^2\theta } \\ =\sqrt{\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }} \\ =\sqrt{\frac{{\sin }^2\theta +{\cos }^2\theta }{{\cos }^2\theta {\sin }^2\theta }} \\ =\sqrt{\frac{1}{{\cos }^2\theta {\sin }^2\theta }} \\ =\frac{1}{\cos \theta \sin \theta } \\ RHS=\tan \theta +cot\theta \\ =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } \\ =\frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta \sin \theta } \\ =\frac{1}{\cos \theta \sin \theta } \\ Therefore,LHS=RHS \\ Henceproved \end{gathered}$$
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