Question

Prove that $\sqrt{(se{c}^2\theta +\cos e{c}^2\theta )}=\tan \theta +cot\theta$

Answer 1

As we have $\sqrt{(se{c}^2\theta +\cos e{c}^2\theta )}=\tan \theta +cot\theta$

Now, taking LHS

$\sqrt{se{c}^2\theta +\cos e{c}^2\theta }$

$Assec\theta =\frac{1}{\cos \theta },\cos ec\theta =\frac{1}{\sin \theta }$

$$\begin{gathered} \Rightarrow \sqrt{\left(\frac{1}{{\cos }^2\mathrm{\theta }}\right)+\left(\frac{1}{{\sin }^2\mathrm{\theta }}\right)} \\ \Rightarrow \sqrt{\left(\frac{{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }}{{\cos }^2{\mathrm{\theta sin}}^2\mathrm{\theta }}\right)} \\ \Rightarrow \sqrt{\frac{1}{{\cos }^2{\mathrm{\theta sin}}^2\mathrm{\theta }}}\left[{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1\right] \\ \Rightarrow \frac{1}{\mathrm{cos\theta sin\theta }} \\ \Rightarrow \frac{{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }}{\mathrm{cos\theta sin\theta }} \\ \Rightarrow \frac{{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta sin\theta }}+\frac{{\cos }^2\mathrm{\theta }}{\mathrm{cos\theta sin\theta }} \\ \Rightarrow \frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }} \\ \Rightarrow \mathrm{tan\theta }+\mathrm{cot\theta } \\ \mathrm{RHS}. \end{gathered}$$

Hence Proved.