Question

If $x=tan[\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}\frac{a}{b}]$ and $y=tan[\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}\frac{a}{b}]$, then

• A. $x+y=2b/a$
• B. $x+y=2a/b$
• C. $x-y=\frac{2\sqrt{b^2-a^2}}{a}$
• D. $x-y=\frac{2\sqrt{a^2-b^2}}{b}$

Answer 1

Answer: (A), (C)

The correct options are
A $x+y=2b/a$
C $x-y=\frac{2\sqrt{b^2-a^2}}{a}$

Let
$\frac{1}{2}co{s}^{-1}\left(\frac{a}{b}\right)=A$
Hence
$x+y$
$=\frac{1+tanA}{1-tanA}+\frac{1-tanA}{1+tanA}$
$=\frac{(1+tanA{)}^2+(1-tanA{)}^2}{1-ta{n}^{2}A}$
$=2\frac{1+ta{n}^{2}A}{1-ta{n}^{2}A}$
$=2sec\left(2A\right)$
$=\frac{2}{cos\left(co{s}^{-1}\right(\frac{a}{b}\left)\right)}$
$=\frac{2b}{a}$
Similarly $x-y$
$=\frac{(1+tanA{)}^2-(1-tanA{)}^2}{1-ta{n}^{2}A}$
$=\frac{2\left(2tanA\right)}{1-ta{n}^{2}A}$
$=2\left(sin\right(2A\left)\right)$
$=2\left(\sqrt{1-co{s}^{2}2A}\right)$
$=2\left(\sqrt{1-\frac{a^2}{b^2}}\right)$
$=\frac{2\sqrt{b^2-a^2}}{b}$