# For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha.\cos^2\alpha+\cos^4\alpha$

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.

• Write the $sin^2$ with cos one as 1-cos^2 Aug 29, 2020 at 12:46
• Try factoring $\sin^2\alpha$ out of the first two terms and $\cos^4\alpha$ out of the last two terms in the numerator. Aug 29, 2020 at 12:46
• @AdityaDwivedi, in the beginning, or to continue my try? Aug 29, 2020 at 12:47
• In the beginning Aug 29, 2020 at 12:47
• I was looking at your final expression when I made that suggestion. See DatBoi's answer. That said, Safdar's answer is also good. Aug 29, 2020 at 12:56

## 3 Answers

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}