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. ${\tan }^2\alpha =\frac{ab(ab-4)}{(a+b{)}^2}$
• D. ${\sec }^2\alpha =\frac{(a-b{)}^2+a^2{b}^2}{(a+b{)}^2}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $ab>4$
C ${\tan }^2\alpha =\frac{ab(ab-4)}{(a+b{)}^2}$
D ${\sec }^2\alpha =\frac{(a-b{)}^2+a^2{b}^2}{(a+b{)}^2}$

$tan\theta +tan\varphi =a$
$cot\theta +cot\varphi =b$
$\frac{tan\theta +tan\varphi }{tan\theta tan\varphi }=b$
$tan\theta tan\varphi =\frac{a}{b}$
Hence $tan\theta -tan\varphi =\sqrt{a^2-4\frac{a}{b}}$
$=\sqrt{\frac{a^{2}b-4a}{b}}$
Hence $tan\alpha =\frac{\sqrt{\frac{a^{2}b-4a}{b}}}{1+\frac{a}{b}}$
$=\frac{\sqrt{a^2{b}^2-4ab}}{b+a}$
Hence $ta{n}^2\alpha =\frac{ab(ab-4)}{(a+b{)}^2}$
Now $se{c}^2\alpha =1+ta{n}^2\alpha$
$\Rightarrow se{c}^2\alpha =\frac{(a-b{)}^2+a^2{b}^2}{(a+b{)}^2}$
Also, since $\frac{ab(ab-4)}{(a+b{)}^2}>0,ab>4$