Question

sin²θ tanθ + cos²θ cot θ + 2sin θ cos θ = tan θ + cot θ

Answer 1

$$\begin{gathered} {\sin }^2\theta \tan \theta +{\cos }^2\theta \cot \theta +2\sin \theta \cos \theta \\ =\frac{{\sin }^3\theta }{\cos \theta }+\frac{{\cos }^3\theta }{\sin \theta }+2\sin \theta \cos \theta \\ =\frac{{\sin }^4\theta +{\cos }^4\theta +2{\sin }^2\theta {\cos }^2\theta }{\sin \theta \cos \theta } \\ =\frac{{\left({\sin }^2\theta +{\cos }^2\theta \right)}^2}{\sin \theta \cos \theta } \end{gathered}$$
$$\begin{gathered} =\frac{1}{\sin \theta \cos \theta }\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta } \\ =\frac{{\sin }^2\theta }{\sin \theta \cos \theta }+\frac{{\cos }^2\theta }{\sin \theta \cos \theta } \end{gathered}$$
$$\begin{gathered} =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } \\ =\tan \theta +\cot \theta \end{gathered}$$