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Problem :

If $\tan^2\alpha \tan^2\beta +\tan^2\beta \tan^2\gamma + \tan^2\gamma \tan^2\alpha + 2\tan^2\alpha \tan^2\beta \tan^2\gamma =1$

Then find the value of $\sin^2\alpha + \sin^2\beta +\sin^2\gamma$

Please suggest how to proceed in such problem.

Sachin
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1 Answers1

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HINT:

If $\sin^2\alpha=x$ etc,

$\displaystyle\tan^2\alpha=\frac{\sin^2\alpha}{\cos^2\alpha}=\frac{\sin^2\alpha}{1-\sin^2\alpha}=\frac x{1-x}$

Just simplify to find $x+y+z=1$ assuming $(1-x)(1-y)(1-z)\ne0$ i.e, $\tan\alpha$ etc. are finite

lab bhattacharjee
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  • $tan^2\alpha tan^2\beta +tan^2\beta tan^2\gamma + tan^2\gamma tan^2\alpha + 2tan^2\alpha tan^2\beta tan^2\gamma =1$ .....(i)

    putting tan$^2\alpha = \frac{x}{1-x}; tan^2\beta = \frac{y}{1-y}; tan^2\beta = \frac{z}{1-z}$

    After simplifying by putting these values in (i) I got :

    $xy +yz +zx -3xyz = (1-x)(1-y)(1-z)$ Now what to do further.. to get x +y +z

     – Sachin Nov 08 '13 at 16:25
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    @sultan, I found $$\sum \frac{xy}{(1-x)(1-y)}=1-\frac{2xyz}{(1-x)(1-y)(1-z)}\implies \cdots x+y+z=1$$ – lab bhattacharjee Nov 08 '13 at 16:28
  • @sultan, do you have any further confusion? – lab bhattacharjee Nov 12 '13 at 08:52