# How am I supposed to expand $\sin^2 A + \sin^4 A = 1$ into $1 + \sin^2A = \tan^2A$?

My question is how can i expand $$\sin^2 A + \sin^4 A = 1$$ into: $$1 + \sin^2A = \tan^2A$$

I tried quite a few ways I know but all of them kinda felt random. i am not sure how to share my trials here. I am quite beginner in trigonometry. it is one of the extra test question from my textbook. I don't need it but cant control curiosity. so pls help me.

EDIT:
found the solution, dropping it here,
\begin{align} \sin^2 A + \sin^4 A & = 1 \\ \sin^4 A & = 1 - sin^2 A \\ \sin^2 A . \sin^2 A & = cos^2 A \\ \sin^2 A . (1 - \cos^2 A) & = cos^2 A \\ \sin^2 A - \sin^2 A.\cos^2 A & = cos^2 A \\ \sin^2 A & = cos^2 A + \sin^2 A.\cos^2 A \\ \sin^2 A & = \cos^2 A(1 + \sin^2 A) \\ 1 + \sin^2 A & = \cfrac{\sin^2 A}{\cos^2 A} \\ 1 + \sin^2 A & = \tan^2 A \\ \end{align}

Hint

If $$\sin^2A + \sin^4A = 1 \implies \sin^4A = \cos^2 A$$

Can you proceed from here?

$$\sin^2A(1-\cos^2A) = \cos^2A$$

$$\sin^2A = \cos^2A + \sin^2A\cos^2A$$

Divide by $$\cos^2 A$$

• i was able to prove this, but dont know how to proceed. pls help Apr 26, 2021 at 10:09
• i got it. Thanks a lot! Apr 26, 2021 at 10:20

$$\sin^2 A = 1 - \sin^4 A$$ $$\sin^2A= (1 - \sin^2 A)(1 + \sin^2 A)$$ $$\sin^2 A= (\cos^2 A) (1 + \sin^2 A)$$ $$\therefore \tan^2 A = 1 + \sin^2 A$$

• Thanks for the edit! It can fit in one line but you showed me that it looks better like this. Apr 26, 2021 at 11:01
• And also nice proof +1 Apr 26, 2021 at 11:01
• Yep, I thought about it for quite a while. It's certainly nicer than the book's solution. Apr 26, 2021 at 11:02
• Nice ........... Apr 26, 2021 at 11:24

Writing $$\sin^2A=a,$$ we have $$a+a^2=1$$

We need $$1+a=\dfrac a{1-a}$$

As $$1-a\ne0,1+1^2\ne1$$ $$\iff a=(1-a)(1+a)\iff a=1-a^2\iff a+a^2=1$$

Done!