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 ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma$ is equal to

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

Rewrite $ta{n}^2\alpha as\frac{si{n}^2\alpha }{1-si{n}^2\alpha }$
Do the same for $\beta and\gamma$
and then simplify.