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

• A. 0
• B. – 1
• C. 1
• D. None of these

Answer 1

Answer: (C)

1

Answer 2

$\sin^{2}\alpha + \sin^{2}\beta + \sin^{2}\gamma$

$= \frac{\tan^{2}\alpha}{1 + \tan^{2}\alpha} + \frac{\tan^{2}\beta}{1 + \tan^{2}\beta} + \frac{\tan^{2}\gamma}{1 + \tan^{2}\gamma}$

$= \frac{x}{1 + x} + \frac{y}{1 + y} + \frac{z}{1 + z}$ $(x = \tan^{2}\alpha,y = \tan^{2}\beta,z = \tan^{2}\gamma)$

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

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

$(\because xy + yz + zx + 2xyz = 1)$