Question

If $tan\theta +tan\varphi =a,cot\theta +cot\varphi =b,\theta -\varphi =\alpha \ne 0$ then

• A. $ab>4$
• B. $ab=4$
• C. $ta{n}^2\alpha =\frac{ab(ab-4)}{(a+b{)}^2}$
• D. $co{t}^2\alpha =\frac{ab(ab+4)}{(a-b{)}^2}$

Answer 1

Answer: (A), (C)

$ta{n}^2\alpha =\frac{ab(ab-4)}{(a+b{)}^2}$
$ab>4$

Given $tan\theta +tan\varphi =a,cot\theta +cot\varphi =b$

$\frac{tan\theta +tan\alpha }{cot\theta +cot\alpha }=\frac{a}{b}\Rightarrow \frac{tan\theta +tan\alpha }{\frac{1}{tan\theta }+\frac{1}{tan\alpha }}=\frac{a}{b}$

$\Rightarrow tan\theta tan\alpha =\frac{a}{b}$ ...(1)

$\alpha =\theta -\varphi \Rightarrow tan\alpha =tan(\theta -\varphi )$

$\Rightarrow ta{n}^2\alpha =ta{n}^2(\theta -\varphi )={\left(\frac{tan\theta -tan\alpha }{1+tan\theta tan\alpha }\right)}^2={\left(\frac{\sqrt{(tan\theta +tan\alpha {)}^2-4tan\theta tan\alpha }}{1+tan\theta tan\alpha }\right)}^2$

From (1)

$ta{n}^2\alpha ={\left(\frac{\sqrt{a^2-4\frac{a}{b}}}{1+\frac{a}{b}}\right)}^2=\frac{ab(ab-4)}{{(a+b)}^2}$

As $ta{n}^2\alpha >0\Rightarrow ab>4$