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

The correct option is C

$1$

Explanation for the correct option:

Step 1: Given that,

${\tan }^2\alpha {\tan }^2\beta +{\tan }^2\beta {\tan }^2\gamma +{\tan }^2\gamma {\tan }^2\alpha +2{\tan }^2\alpha {\tan }^2\beta {\tan }^2\gamma =1....\left(1\right)$

Let $\begin{array}{rcl}{\sin }^2\alpha & =& x\end{array}$

Then, $\begin{array}{rcl}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\alpha }& \mathbf{=}& \mathbf{1}\mathbf{–}\mathit{x}\mathrm{and}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\mathit{\alpha }\mathbf{=}\frac{\mathbf{x}}{\mathbf{1}\mathbf{-}\mathbf{x}}\end{array}$

Similarly, let ${\sin }^2\beta =y$

Then, ${\cos }^2\beta =1–y\mathbf{}\mathbf{and}{\tan }^2\beta =\frac{y}{1-y}$

And, let ${\sin }^2\gamma =z$

Then, ${\cos }^2\gamma =1–z\mathbf{and}\mathbf{}{\tan }^2\gamma =\frac{z}{1-z}$

Step 2: Substituting these values in $\left(1\right)$,

$\begin{array}{rcl}\therefore \left(\frac{x}{1-x}\right)\left(\frac{y}{1-y}\right)+\left(\frac{y}{1-y}\right)\left(\frac{z}{1-z}\right)+\left(\frac{z}{1-z}\right)\left(\frac{x}{1-x}\right)+2\left(\frac{x}{1-x}\right)\left(\frac{y}{1-y}\right)\left(\frac{z}{1-z}\right)& =& 1\\ & \Rightarrow & \left[\frac{xy}{\left(1-x\right)\left(1-y\right)}\right]+\left[\frac{yz}{\left(1-y\right)\left(1-z\right)}\right]+\left[\frac{zx}{\left(1-z\right)\left(1-x\right)}\right]+2\left[\frac{xyz}{\left(1-x\right)\left(1-y\right)\left(1-z\right)}\right]=1\\ & \Rightarrow & \left[\frac{xy\left(1-z\right)+yz\left(1-x\right)+zx\left(1-y\right)}{\left(1-x\right)\left(1-y\right)\left(1-z\right)}\right]=1-2\left[\frac{xyz}{\left(1-x\right)\left(1-y\right)\left(1-z\right)}\right]\\ & \Rightarrow & xy\left(1-z\right)+yz\left(1-x\right)+zx\left(1-y\right)=\left(1-x\right)\left(1-y\right)\left(1-z\right)-2xyz\\ \therefore x+y+z& =& 1\end{array}$

Therefore, ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =1$

Hence, the correct option is (C).