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.
Best Answer
You're almost there! $$\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\overbrace{\sin^2\alpha\cdot\cos^4\alpha+\cos^6\alpha}}{\cos^2\alpha}$$ $$=\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\cos^4(\alpha)(\cos^2(\alpha)+\sin^2(\alpha))}{\cos^2\alpha}$$ $$=\dfrac{\sin^2\alpha+ \overbrace{\sin^2\alpha\cdot\cos^2\alpha+\cos^4(\alpha)}}{\cos^2\alpha}$$ $$=\dfrac{\sin^2\alpha+\cos^2(\alpha)(\cos^2(\alpha)+\sin^2(\alpha))}{\cos^2\alpha}$$ $$=\dfrac{\overbrace{\sin^2\alpha+\cos^2(\alpha)}}{\cos^2\alpha}$$ $$=\dfrac{1}{\cos^2\alpha}$$ $$=\boxed{\sec^2(\alpha)}$$