The value of tan 9° – tan 27° – tan 63° + tan 810, is

The value of tan 9° - tan 27° - tan 63° + tan 810, is

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Solution:

We have,

$\begin{array}{l}\mathrm{tan}{9}^{\circ }-\mathrm{tan}{27}^{\circ }-\mathrm{tan}{63}^{\circ }+\mathrm{tan}{81}^{\circ }\\ =\left(\mathrm{tan}{9}^{\circ }+\mathrm{tan}{81}^{\circ }\right)-\left(\mathrm{tan}{27}^{\circ }+\mathrm{tan}{63}^{\circ }\right)\\ =\frac{1}{\mathrm{cos}{9}^{\circ }\mathrm{cos}{81}^{\circ }}-\frac{1}{\mathrm{cos}{27}^{\circ }\mathrm{cos}{63}^{\circ }}\\ =\frac{1}{\mathrm{sin}{9}^{\circ }\mathrm{cos}{9}^{\circ }}-\frac{1}{\mathrm{sin}{27}^{\circ }\mathrm{cos}{27}^{\circ }}\\ =\frac{2}{\mathrm{sin}{18}^{\circ }}-\frac{2}{\mathrm{sin}{54}^{\circ }}\\ =2\left\{\frac{\mathrm{sin}54-\mathrm{sin}{18}^{\circ }}{\mathrm{sin}{54}^{\circ }\mathrm{sin}{18}^{\circ }}\right\}=2\left\{\frac{2\mathrm{cos}{36}^{\circ }\mathrm{sin}{18}^{\circ }}{\mathrm{sin}{18}^{\circ }\mathrm{cos}{36}^{\circ }}\right\}=4\end{array}$

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