Question

sec square(90-θ) - cot squareθ/ 2(sin square 25 + sin square 65) - 2 cos square 60 tan square 28 tan square 62/ 3(sec square 43 - cot square 47). evaluate this

Answer 1

$$\begin{gathered} \frac{{\sec }^2\left(90°-\mathrm{\theta }\right)-{\cot }^2\mathrm{\theta }}{2\left[{\sin }^{2}25°+{\sin }^{2}65°\right]}-\frac{2{\cos }^{2}60°{\tan }^{2}28°{\tan }^{2}62°}{3\left[{\sec }^{2}43°-{\cot }^{2}47°\right]} \\ =\frac{{\mathrm{cosec}}^2\mathrm{\theta }-{\cot }^2\mathrm{\theta }}{2\left[{\sin }^{2}25°+{\cos }^2\left(90°-25°\right)\right]}-\frac{2\times {\left(1/2\right)}^2\times {\tan }^2\left(90°-62°\right)\times {\tan }^{2}62°}{3\left[{\sec }^{2}43°-{\cot }^2\left(90°-43°\right)\right]} \\ =\frac{1}{2\left({\sin }^{2}25°+{\cos }^{2}25°\right)}-\frac{1\times {\cot }^{2}62°\times {\tan }^{2}62°}{6\left[{\sec }^{2}43°-{\tan }^{2}43°\right]} \\ =\frac{1}{2\times 1}-\frac{{\displaystyle \frac{1}{{\tan }^{2}62°}}\times {\tan }^{2}62°}{6\times 1} \\ =\frac{1}{2}-\frac{1}{6} \\ =\frac{3-1}{6} \\ =\frac{2}{6} \\ =\frac{1}{3} \end{gathered}$$