Question

$\frac{{\sec }^2\mathrm{\theta }-{\cot }^2(90°-\mathrm{\theta })}{{\mathrm{cosec}}^{2}67°-{\tan }^{2}23°}+({\sin }^{2}40°+{\sin }^{2}50°)$

Answer 1

$$\begin{gathered} \begin{array}{l}\frac{{\sec }^2\theta -{\cot }^2({90}^0-\theta )}{\cos {\text{ec}}^2{67}^0-{\tan }^2{23}^0}+({\sin }^2{40}^0+{\sin }^2{50}^0)\\ \end{array} \\ \begin{array}{l}\text{=}\frac{{\sec }^2\theta -{\tan }^2\theta }{\cos {\text{ec}}^2{67}^0-{\tan }^2({90}^0-{67}^0)}+{\sin }^2{40}^0+{\sin }^2({90}^0-{40}^0)\\ \end{array} \\ \begin{array}{l}\text{=}\frac{1}{\cos {\text{ec}}^2{67}^0-{\cot }^2{67}^0}+({\sin }^2{40}^0+{\cos }^2{40}^0)\\ \end{array} \\ \begin{array}{l}\text{=1+1}\\ \end{array} \\ \text{=2} \end{gathered}$$