Question

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 the value of ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma$

• A. $0$
• B. $-1$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $1$

$weknow{\tan }^2\alpha =\frac{{\sin }^2\alpha }{{\cos }^2\alpha }=\frac{{\sin }^2\alpha }{1-{\sin }^2\alpha }$
$Let{\sin }^2\alpha =x$
${\tan }^2\alpha =\frac{x}{1-x}$
$similarlyif{\sin }^2\beta =y,{\tan }^2\beta =\frac{y}{1-y}$
$similarlyif{\sin }^2\gamma =z,{\tan }^2\gamma =\frac{z}{1-z}$
$putting\text{these values in the given equation}$
$\text{2}\left(\frac{x}{1-x}\right)\left(\frac{y}{1-y}\right)\left(\frac{z}{1-z}\right)+\left(\frac{x}{1-x}\right)\left(\frac{y}{1-y}\right)+\left(\frac{y}{1-y}\right)\left(\frac{z}{1-z}\right)+\left(\frac{z}{1-z}\right)\left(\frac{x}{1-x}\right)=1$
$\Rightarrow \frac{xy}{(1-y)(1-x)}+\frac{yz}{(1-y)(1-z)}+\frac{zx}{(1-z)(1-x)}=1-\frac{2xyz}{(1-x)(1-y)(1-z)}$
$\Rightarrow \frac{xy-xyz+yz-xyz+zx-xyz}{(1-x)(1-y)(1-z)}=1-\frac{2xyz}{(1-x)(1-y)(1-z)}$
$\Rightarrow (x+y+z)-\frac{2xyz}{(1-x)(1-y)(1-z)}=1-\frac{2xyz}{(1-x)(1-y)(1-z)}$
$\Rightarrow x+y+z=1$
$Hence$
${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =1$