Question

lf ${\tan }^2\alpha {\tan }^2\beta +{\tan }^2\beta {\tan }^2\alpha +{\tan }^2\gamma {\tan }^2\alpha +2{\tan }^2\alpha {\tan }^2\beta {\tan }^2\gamma =1$, then ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =$

• A. $0$
• B. $-1$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

Let $ta{n}^2\alpha =x$

$ta{n}^2\beta =y$

$ta{n}^2\gamma =z$

Hence

$xy+yz+zx+2xyz=1$ ...(i)

Now

$tan\alpha =\sqrt{x}$

$1+ta{n}^2\alpha =1+x$

${\sec }^2\alpha =1+x$

${\sin }^2\alpha =1-\frac{1}{1+x}$

$sin\alpha =\sqrt{\frac{x}{1+x}}$

Hence

$si{n}^2\alpha =\frac{x}{1+x}$

Similarly

$si{n}^2\beta =\frac{y}{1+y}$

$si{n}^2\gamma =\frac{z}{1+z}$

Hence

$si{n}^2\alpha +si{n}^2\beta +si{n}^2\gamma$

$=\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}$

$=\frac{x+y+2xy}{1+(x+y)+xy}+\frac{z}{1+z}$

$=\frac{x+y+z+2xy+zx+zy+2xyz+zx+zy+xyz}{1+z+x+y+z(x+y)+xy+z\left(xy\right)}$

$=\frac{x+y+z+(xy+yz+zx+2xyz)+(xy+yz+zx+xyz)}{1+x+y+z+(xy+yz+zx+xyz)}$

now

$xy+yz+zx+2xyz=1$

Hence

Substituting, we get

$\frac{1+x+y+z+(xy+yz+zx+xyz)}{1+x+y+z+(xy+yz+zx+xyz)}$

$=1$