Bab 16 Fungsi Trigonometri

$\begin{array}{l}\text{Sebelah kanan }\frac{{\tan }^{2}x-1}{{\tan }^{2}x+1}\\ =\frac{\frac{{\sin }^{2}x}{{\cos }^{2}x}-1}{\frac{{\sin }^{2}x}{{\cos }^{2}x}+1}\leftarrow \boxed{\tan x=\frac{\sin x}{\cos x}}\\ =\frac{\frac{{\sin }^{2}x-{\cos }^{2}x}{{\cos }^{2}x}}{\frac{{\sin }^{2}x+{\cos }^{2}x}{{\cos }^{2}x}}\\ =\frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x+{\cos }^{2}x}\\ ={\sin }^{2}x-{\cos }^{2}x\leftarrow \boxed{{\sin }^{2}x+{\cos }^{2}x=1}\\ =\text{Sebelah kiri}\end{array}$

$\begin{array}{l}\text{Sebelah kiri}\\ ={\tan }^2\theta -{\sin }^2\theta \\ =\frac{{\sin }^2\theta }{{\cos }^2\theta }-{\sin }^2\theta \\ =\frac{{\sin }^2\theta -{\sin }^2\theta {\cos }^2\theta }{{\cos }^2\theta }\\ =\frac{{\sin }^2\theta \left(1-{\cos }^2\theta \right)}{{\cos }^2\theta }\\ =\frac{{\sin }^2\theta {\sin }^2\theta }{{\cos }^2\theta }\\ =\left(\frac{{\sin }^2\theta }{{\cos }^2\theta }\right)\left({\sin }^2\theta \right)\\ ={\tan }^2\theta {\sin }^2\theta \\ =\text{Sebelah kanan}\end{array}$