Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identities
tan x = [tex]\frac{sinx}{cosx}[/tex] , sin²x + cos²x = 1
Consider the left side
[tex]\frac{1-tan^2x}{1+tan^2x}[/tex]
= [tex]\frac{1-\frac{sin^2x}{cos^2x} }{1+\frac{sin^2x}{cos^2x} }[/tex] ← multiply numerator/ denominator of the whole fraction by cos²x
= [tex]\frac{cos^2x-sin^2x}{cos^2x+sin^2x}[/tex]
= [tex]\frac{cos^2x-sin^2x}{1}[/tex]
= cos²x - sin²x
= cos²x - (1 - cos²x)
= cos²x - 1 + cos²x
= 2cos²x - 1
= right side , thus proven
[tex]\frac{1 - \tan^2(x)}{1 + \tan^2(x)}\\= \frac{\frac{\cos^2(x)}{\cos^2(x)} - \frac{\sin^2(x)}{\cos^2(x)}}{\frac{\cos^2(x)}{\cos^2(x)} + \frac{\sin^2(x)}{\cos^2(x)}}\\= \frac{\frac{\cos^2(x) - \sin^2(x)}{\cos^2(x)}}{\frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}}\\= \frac{\cos^2(x) - \sin^2(x)}{\cos^2(x)} \cdot \frac{\cos^2(x)}{\cos^2(x) + \sin^2(x)}\\= \frac{(\cos^2(x) - \sin^2(x)) \cdot \cos^2(x)}{\cos^2(x) \cdot (\cos^2(x) + \sin^2(x))}\\= \frac{\cos^2(x) - \sin^2(x)}{\cos^2(x) + \sin^2(x)}[/tex]
[tex]= \cos^2(x) - \sin^2(x)\\= \cos^2(x) - (1 - \cos^2(x))\\= \cos^2(x) - 1 + \cos^2(x)\\= 2\cos^2(x) - 1[/tex]
