Question

$\frac{{\cos }^{2}56°+{\cos }^{2}34°}{{\sin }^{2}56°+{\sin }^{2}34°}+3{\tan }^{2}56°{\tan }^{2}34°=?$
(a) $3\frac{1}{2}$
(b) 4
(c) 6
(d) 5

Answer 1

(b) 4

$$\begin{gathered} \frac{{\cos }^2{56}^0+{\cos }^2{34}^0}{{\sin }^2{56}^0+{\sin }^2{34}^0}+3{\text{tan}}^2{56}^0{\text{tan}}^2{34}^0 \\ \text{=}\frac{{\left\{\cos ({90}^0-{34}^0)\right\}}^2+{\cos }^2{34}^0}{{\left\{\sin ({90}^0-{34}^0)\right\}}^2+{\sin }^2{34}^0}+3{\left\{\tan ({90}^0-{34}^0)\right\}}^2{\text{tan}}^2{34}^0 \\ \text{=}\frac{{\sin }^2{34}^0+{\cos }^2{34}^0}{{\cos }^2{34}^0+{\sin }^2{34}^0}+3{\cot }^2{34}^0{\text{tan}}^2{34}^0\left[\begin{array}{l}\because \cos ({90}^0-\theta )=\sin \theta ,\sin ({90}^0-\theta )\\ =\cos \theta \text{and}\tan ({90}^0-\theta )=\cot \theta \end{array}\right] \\ \text{=}\frac{1}{1}+3\times 1[\because \cot \theta =\frac{1}{\tan \theta }\text{and}{\sin }^2\theta +{\cos }^2\theta =1] \\ \text{=4} \end{gathered}$$