Question

If $\frac{\sin x}{a}=\frac{\cos x}{b}=\frac{\tan x}{c}=k,$ then $bc+\frac{1}{ck}+\frac{ak}{1+bk}=$

• A. $k\left(a+\frac{1}{a}\right)$
• B. $\frac{1}{k}\left(a+\frac{1}{a}\right)$
• C. $\frac{1}{k^2}$
• D. $\frac{a}{k}$

Answer 1

Answer: (B)

$bc+\frac{1}{ck}+\frac{ak}{1+bk}$
$=\left(\frac{\cos x}{k}\right)\left(\frac{\tan x}{k}\right)+\frac{1}{\tan x}+\frac{\sin x}{1+\cos x}$
$=\frac{\sin x}{k^2}+\left(\frac{\cos x}{\sin x}+\frac{\sin x}{1+\cos x}\right)$
$=\frac{1}{k}\left(\frac{\sin x}{k}\right)+\frac{\cos x\left(1+\cos x\right)+{\sin }^{2}x}{\sin x\left(1+\cos x\right)}$
$=\frac{a}{k}+\frac{1+\cos x}{\sin x\left(1+\cos x\right)}$
$=\frac{a}{k}+\frac{1}{\sin x}=\frac{a}{k}+\frac{1}{ak}=\frac{1}{k}\left(a+\frac{1}{a}\right)$