Question

Given $\frac{sinx}{a}=\frac{cosx}{b}=\frac{tanx}{c}=k$

Answer 1

As $\frac{\sin x}{a}=\frac{\cos x}{b}=\frac{\tan x}{c}=k$
A) $a^2+b^2=\frac{1}{k^2}({\sin }^{2}x+{\cos }^{2}x)=\frac{1}{k^2}$
B) $a^2+b^2+c^2=\frac{1}{k^2}({\sin }^{2}x+{\cos }^{2}x+{\tan }^{2}x)$
$=\frac{1+{\tan }^{2}x}{k^2}=\frac{{\sec }^{2}x}{k^2}=\frac{1}{b^2{k}^4}$
C) $bc=\frac{\cos x\tan x}{k^2}=\frac{\sin x}{k^2}=\frac{ak}{k^2}=\frac{a}{k}$
D) $\frac{1}{ck}+\frac{ak}{1+bk}=\frac{1}{\tan x}+\frac{\sin x}{1+\cos x}$
$=\frac{\cos x(1+\cos x)+{\sin }^{2}x}{\sin x(1+\cos x)}=\frac{1}{\sin x}=\frac{1}{ak}$