Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identities
1 + tan²A = sec²A , secA = [tex]\frac{1}{cosA}[/tex], cosecA = [tex]\frac{1}{sinA}[/tex] , cotA = [tex]\frac{cosA}{sinA}[/tex] , tanA = [tex]\frac{sinA}{cosA}[/tex]
Consider the left side
[tex]\frac{(1+tan^2A)cotA}{cosec^2A}[/tex]
= [tex]\frac{sec^2AcotA}{cosec^2A}[/tex]
= [tex]\frac{\frac{1}{cos^2A}(\frac{cosA}{sinA}) }{cosec^2A}[/tex]
= [tex]\frac{\frac{1}{cosAsinA} }{\frac{1}{sin^2A} }[/tex]
= [tex]\frac{1}{cosAsinA}[/tex] × sin²A
= [tex]\frac{sinA}{cosA}[/tex]
= tanA = right side , thus proven
