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If $2\tan^2A\tan^2B\tan^2C + \tan^2A\tan^2B + \tan^2B\tan^2C + \tan^2C\tan^2A = 1$, then find the value of $\sin^2A + \sin^2B + \sin^2C$.

My attempt

1). I tried to multiply both sides by $\cos^2A\cos^2B\cos^2C$ in given. Then took common but after that, I didn't get what to do next

2). I tried to relate the given with $(ab+bc+ca)^2,$ but it also didn't work... I could only think of these two ways but nothing worked for me...

YuiTo Cheng
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Kenji
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1 Answers1

2

$$\sum_{cyc}\sin^2\alpha=\sum_{cyc}\frac{\tan^2\alpha}{1+\tan^2\alpha}=\frac{\sum\limits_{cyc}\tan^2\alpha(1+\tan^2\beta)(1+\tan^2\gamma)}{\prod\limits_{cyc}(1+\tan^2\alpha)}=$$ $$=\frac{\sum\limits_{cyc}(\tan^2\alpha\tan^2\beta\tan^2\gamma+2\tan^2\alpha\tan^2\beta+\tan^2\alpha)}{\prod\limits_{cyc}(1+\tan^2\alpha)}=$$ $$=\frac{\tan^2\alpha\tan^2\beta\tan^2\gamma+1+\sum\limits_{cyc}(\tan^2\alpha\tan^2\beta+\tan^2\alpha)}{\prod\limits_{cyc}(1+\tan^2\alpha)}=1.$$

Michael Rozenberg
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  • Thanks for answering.. But I'm in 12th grade, I have not seen this type of solving method, if you have time and know any simpler method, I will be a great help.. Again Thanks bro – Kenji Jun 15 '19 at 02:00
  • @Kenji I think my solution is very easy and smooth. I am ready to explain. Which step is not clear? – Michael Rozenberg Jun 15 '19 at 02:03
  • I didn't understand why you used the subscript "cyc" in summation sign and Multiplication sign. This is the first I saw this method of solving question.. I am used to summation goes from 1 to n. – Kenji Jun 15 '19 at 02:16
  • @Kenji it's a designation only for to write less. For example, $\sum\limits_{cyc}\sin^2\alpha=\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ or $\prod\limits_{cyc}(1+\tan^2\alpha)=(1+\tan^2\alpha)(1+\tan^2\beta)(1+\tan^2\gamma).$ It's nothing! – Michael Rozenberg Jun 15 '19 at 02:27
  • Thanks ..now I got it . – Kenji Jun 15 '19 at 02:41
  • @Kenji You are welcome! – Michael Rozenberg Jun 15 '19 at 02:59
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    Nice solution.... – DXT Jun 15 '19 at 06:00