8. Buktikan bahwa: d. cos 2 θ = 1 + tan 2 θ 1 − tan 2 θ ​

terbukti jika $$\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$$. 

Ingat kembali:

• rumus cosinus untuk sudut ganda: $$\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$$  
• identitas trigonometri: $${\tan }^2\alpha +1=se{c}^2\alpha$$ 
• $$\tan \alpha =\frac{1}{\text{sec}\alpha }$$ 
• $$\cos \alpha =\frac{1}{\text{sec}\alpha }$$

Oleh karena itu, dapat diperoleh:

 $$\begin{array}{rcl}\cos 2\theta & =& {\cos }^2\theta -{\sin }^2\theta \\ & =& \left({\cos }^2\theta -{\sin }^2\theta \right)\cdot 1\\ & =& \left({\cos }^2\theta -{\sin }^2\theta \right)\cdot \left(\frac{1}{{\cos }^2\theta }\cdot \frac{{\cos }^2\theta }{1}\right)\\ & =& \left(\frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta }\right)\cdot {\cos }^2\theta \\ & =& \left(\frac{{\cos }^2\theta }{{\cos }^2\theta }-\frac{{\sin }^2\theta }{{\cos }^2\theta }\right)\cdot {\cos }^2\theta \\ & =& \left(\frac{{\xcancel{{\cos }^2\theta }}^1}{{\xcancel{{\cos }^2\theta }}^1}-\frac{{\sin }^2\theta }{{\cos }^2\theta }\right)\cdot {\cos }^2\theta \\ & =& \left(\frac{1}{1}-\left(\frac{\sin \theta }{\cos \theta }\cdot \frac{\sin \theta }{\cos \theta }\right)\right)\cdot \cos \theta \cdot \cos \theta \\ & =& \left(1-\left(\tan \theta \cdot \tan \theta \right)\right)\cdot \frac{1}{\text{sec }\theta }\cdot \frac{1}{\text{sec }\theta }\\ & =& \left(1-{\tan }^2\theta \right)\cdot \frac{1}{{\text{sec}}^2\theta }\\ & =& \left(1-{\tan }^2\theta \right)\cdot \frac{1}{1+{\tan }^2\theta }\\ & =& \frac{\left(1-{\tan }^2\theta \right)\cdot 1}{1+{\tan }^2\theta }\\ & =& \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\end{array}$$ 

Dengan demikian, terbukti jika $$\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$$.