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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.

Thanks in advance!

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}

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3 Answers 3

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Hint

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

Can you proceed from here?

Additional Info

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

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

Divide by $\cos^2 A$

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  • $\begingroup$ i was able to prove this, but dont know how to proceed. pls help $\endgroup$
    – asierx
    Apr 26, 2021 at 10:09
  • $\begingroup$ i got it. Thanks a lot! $\endgroup$
    – asierx
    Apr 26, 2021 at 10:20
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$$\sin^2 A = 1 - \sin^4 A$$ $$ \sin^2A= (1 - \sin^2 A)(1 + \sin^2 A) $$ $$\sin^2 A= (\cos^2 A) (1 + \sin^2 A)$$ $$\therefore \tan^2 A = 1 + \sin^2 A$$

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

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