Question

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

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

Answer 1

The correct option is B $\frac{a}{k}+\frac{1}{ak}$

$sinx=ak$
$cosx=bk$
$tanx=ck$
Since,$tanx=\frac{sinx}{cosx}$,

$ck=\frac{a}{b}$ Or, $bc=\frac{a}{k}$
$ck=tanx$
$\Rightarrow \frac{1}{ck}=cotx$
$\Rightarrow \frac{ak}{1+bk}=\frac{sinx}{1+cosx}$
$\Rightarrow \frac{1}{ck}+\frac{ak}{1+bk}=cotx+\frac{sinx}{1+cosx}$
$=\frac{cosx+{cos}^{2}x+{sin}^{2}x}{sinx(1+cosx)}$
$=\frac{1+cosx}{sinx(1+cosx)}$
$=cosecx$
Hence, $bc+\frac{1}{ck}+\frac{ak}{1+bk}$ = $\frac{a}{k}+\frac{1}{ak}$