Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identities
tan²x + 1 = sec²x and sec x = [tex]\frac{1}{cosx}[/tex]
Consider the left side
Express the sum as a single fraction
[tex]\frac{secx+tanx+secx-tanx}{(secx-tanx)(secx+tanx)}[/tex]
= [tex]\frac{2secx}{sec^2x-tan^2x}[/tex]
= [tex]\frac{2secx}{tan^2x+1-tan^2x}[/tex]
= 2sec x
= [tex]\frac{2}{cosx}[/tex] = right side ⇒ proven
