Question

Evaluate: $2\ast \left(\frac{{\cos }^2{20}^0+{\cos }^2{70}^0}{{\sin }^2{25}^0+{\sin }^2{65}^0}\right)-\tan {45}^0+\tan {13}^0\tan {23}^0\tan {30}^0\tan {67}^0\tan {77}^0$.

Answer 1

=$2\left[\frac{{\cos }^2({90}^0-{70}^0)+{\cos }^2{70}^0}{{\sin }^2{25}^0+{\sin }^2({90}^0-{25}^0)}\right]-\tan {45}^0+\tan {13}^0\tan {77}^0\tan {23}^0\tan {67}^0\tan {30}^0$

=$2\left[\frac{{\sin }^2{70}^0+{\cos }^2{70}^0}{{\sin }^2{25}^0+{\cos }^2{25}^0}\right]-1+\tan {13}^0\tan ({90}^0-{13}^0)\tan {23}^0\tan ({90}^0-{23}^0)\ast \frac{1}{\sqrt{3}}$

=$2\left(\frac{1}{1}\right)-1+\tan {13}^0\cot {13}^0\tan {23}^0\cot {23}^0\ast \frac{1}{\sqrt{3}}$

=$2-1+1\ast 1\ast \frac{1}{\sqrt{3}}$

=$2-1+\frac{1}{\sqrt{3}}=1+\frac{1}{\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{3}}$

=$\frac{(\sqrt{3}+1)\sqrt{3}}{\sqrt{3}\ast \sqrt{3}}=\frac{3+\sqrt{3}}{3}$.