For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha.$
My try: $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\sin^2\alpha+\dfrac{\sin^2\alpha}{\cos^2\alpha}+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\sin^2\alpha\cdot\cos^4\alpha+\cos^6\alpha}{\cos^2\alpha}.$
This doesn't seem to help much.
You're almost there! $$\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\overbrace{\sin^2\alpha\cdot\cos^4\alpha+\cos^6\alpha}}{\cos^2\alpha}$$ $$=\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\cos^4(\alpha)(\cos^2(\alpha)+\sin^2(\alpha))}{\cos^2\alpha}$$ $$=\dfrac{\sin^2\alpha+ \overbrace{\sin^2\alpha\cdot\cos^2\alpha+\cos^4(\alpha)}}{\cos^2\alpha}$$ $$=\dfrac{\sin^2\alpha+\cos^2(\alpha)(\cos^2(\alpha)+\sin^2(\alpha))}{\cos^2\alpha}$$ $$=\dfrac{\overbrace{\sin^2\alpha+\cos^2(\alpha)}}{\cos^2\alpha}$$ $$=\dfrac{1}{\cos^2\alpha}$$ $$=\boxed{\sec^2(\alpha)}$$
Express everything in terms of $ \cos \alpha=c $ $$ (1-c^2) + (1-c^2)/c^2 + (1-c^2) c^2 + c^4$$
$$= \dfrac{c^2-c^4+1-c^2+c^4-c^6+c^6}{c^2} = \dfrac{1}{c^2}$$
Instead of dividing by $\cos^2\alpha$, you could do the following,
\begin{align}&\sin^2 \alpha + \tan^2 \alpha +\sin^2\alpha +\sin^2\alpha\cos^2\alpha + \cos^4\alpha\\ &= \sin^2 \alpha + \tan^2 \alpha +\sin^2\alpha +\cos^2\alpha(\sin^2\alpha + \cos^2\alpha) \\ &= \sin^2 \alpha + \tan^2 \alpha + \cos^2\alpha \\ &=1+\tan^2\alpha \\ &=\sec^2\alpha \end{align}
