Can someone help me answer 61 and 63?

Answer:
see explanation
Step-by-step explanation:
using the trigonometric identities
• sin²x + cos²x = 1
• tan²x + 1 = sec²x
• cotx = [tex]\frac{1}{tanx}[/tex] and secx = [tex]\frac{1}{cosx}[/tex]
(61)
subtracting the 2 fractions
= [tex]\frac{tan^2x-sec^2x}{tanx}[/tex]
= [tex]\frac{sec^2x-1-sec^2x}{tanx}[/tex]
= [tex]\frac{-1}{tanx}[/tex]
= - cotx
(63)
adding the 2 fractions
= [tex]\frac{cos^2x+(1+sinx)^2}{cosx(1+sinx)}[/tex]
= [tex]\frac{cos^2x+1+2sinx+sin^2x}{cosx(1+sinx)}[/tex]
= [tex]\frac{2(1+sinx)}{cosx(1+sinx)}[/tex]
= [tex]\frac{2}{cosx}[/tex] = 2secx