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 equal to

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

Answer 1

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

$bc+\frac{1}{ck}+\frac{ak}{1+bk}$
Substituting values from
$\frac{sinx}{a}=\frac{cosx}{b}=\frac{tanx}{c}=k$
we get
$\frac{cosx}{k}.\frac{tanx}{k}+\frac{k}{tanxk}+\frac{\frac{sinx}{k}k}{1+\frac{cosx}{k}k}=\frac{sinx}{k^2}+\frac{1}{tanx}+\frac{sinx}{1+cosx}=\frac{sinx}{k^2}+\frac{cosx}{sinx}+\frac{sinx}{1+cosx}=\frac{sinx}{k^2}+\frac{1}{sinx}=\frac{ak}{k^2}+\frac{1}{ak}=\frac{1}{k}(a+\frac{1}{a})$