Question

cot(90-tita)sin(90-tita)/ sin tita + cot40/tan50 - (cos square20 + cos square70). evaluate

Answer 1

$$\begin{gathered} \frac{\cot \left(90°-\mathrm{\theta }\right)\sin \left(90°-\mathrm{\theta }\right)}{\sin \mathrm{\theta }}+\frac{\cot 40°}{\tan 50°}-\left[{\cos }^{2}20°+{\cos }^{2}70°\right] \\ =\frac{\tan \mathrm{\theta }.\cos \mathrm{\theta }}{\sin \mathrm{\theta }}+\frac{\cot \left(90°-50°\right)}{\tan 50°}-\left[{\cos }^2\left(90°-70°\right)+{\cos }^{2}70°\right] \\ =\frac{\sin \mathrm{\theta }.\cos \mathrm{\theta }}{\sin \mathrm{\theta }.\cos \mathrm{\theta }}+\frac{\tan 50°}{\tan 50°}-\left[{\sin }^{2}70°+{\cos }^{2}70°\right] \\ =1+1+1 \\ =3 \end{gathered}$$