Question

$\tan \left[ta{n}^{-1}\right(\frac{1}{a+b})+ta{n}^{-1}(\frac{b}{a^2+ab+1}\left)\right]=$

• A. $a$
• B. $1/a$
• C. $b$
• D. $1/b$

Answer 1

Answer: (B)

$1/a$

$tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$ [formula ]
$A=ta{n}^{-1}\left(\frac{1}{a+b}\right)B=ta{n}^{-1}\left(\frac{b}{a^2+ab+1}\right)$
$\therefore tan(A+B)=\frac{\frac{1}{a+b}+\frac{b}{a^2+ab+1}}{1-\frac{b}{(a+b)(a^2+ab+1)}}[tanta{n}^{-1}x=x]$

$=\frac{a^2+ab+1+b^2+ab}{a^3+a^{2}b+a+b{a}^2+a{b}^2+b-b}$
$=\frac{a^2+b^2+2ab+1}{a^3+2a^{2}b+a{b}^2+a}$
$=\frac{(a^2+b^2+2ab+1)}{a(a^2+b^2+2ab+1)}$
$=\frac{1}{a}$