Question

Assertion :If $\frac{x}{\tan (\theta +\alpha )}=\frac{y}{\tan (\theta +\beta )}=\frac{z}{\tan (\theta +\gamma )}$, then $\sum \frac{x+y}{x-y}{\sin }^2(\alpha -\beta )=0$ Reason: $\frac{x+y}{x-y}=\frac{\tan (\theta +\alpha )+\tan (\theta +\beta )}{\tan (\theta +\alpha )-\tan (\theta +\beta )}=\frac{\sin (2\theta +\alpha +\beta )}{\sin (\alpha +\beta )}$

• A. Both the statements are TRUE and reason is the correct explanation of assertion
• B. Both the statements are TRUE and reason is NOT the correct explanation of assertion
• C. Assertion is TRUE and reason is FALSE
• D. Assertion is FALSE and reason is TRUE

Answer 1

The correct option is C Assertion is TRUE and reason is FALSE

Assertion: $\frac{x}{\tan (\theta +\alpha )}=\frac{y}{\tan (\theta +\beta )}=\frac{z}{\tan (\theta +\gamma )}$

$\Rightarrow \frac{x}{y}=\frac{\tan (\theta +\alpha )}{\tan (\theta +\beta )}=\frac{\sin (\theta +\alpha )\cos (\theta +\beta )}{\cos (\theta +\alpha )\sin (\theta +\beta )}$

Appyling componendo and dividendo, we get

$\frac{x+y}{x-y}=\frac{\sin (\theta +\alpha )\cos (\theta +\beta )+\cos (\theta +\alpha )\sin (\theta +\beta )}{\sin (\theta +\alpha )\cos (\theta +\beta )-\cos (\theta +\alpha )\sin (\theta +\beta )}$

$=\frac{\sin (2\theta +\alpha +\beta )}{\sin (\alpha -\beta )}$

$\therefore$ Clearly reason is incorrect

Now $\frac{x+y}{x-y}\sin (\alpha -\beta )=\sin (2\theta +\alpha +\beta )$

$\Rightarrow \frac{x+y}{x-y}\sin (\alpha -\beta )\sin (\alpha -\beta )=\sin (2\theta +\alpha +\beta )\sin (\alpha -\beta )$

$\Rightarrow \frac{x+y}{x-y}{\sin }^2(\alpha -\beta )=\sin (2\theta +\alpha +\beta )\sin (\alpha -\beta )$

$=\frac{1}{2}\left[\cos \right(2\theta +2\beta )-\cos (2\theta +2\alpha \left)\right]$

Similarly $\Rightarrow \frac{y+z}{y-z}{\sin }^2(\beta -\gamma )=\frac{1}{2}\left[\cos \right(2\theta +2\gamma )-\cos (2\theta +2\beta \left)\right]$

and $\Rightarrow \frac{z+x}{z-x}{\sin }^2(\gamma -\alpha )=\frac{1}{2}\left[\cos \right(2\theta +2\alpha )-\cos (2\theta +2\gamma \left)\right]$

$\therefore \sum \frac{x+y}{x-y}{\sin }^2(\alpha -\beta )=0$

Hence assertion is correct.