1+tan^2A/1+cot^2 is equal to

Question

\(\frac{1+tan^2A}{1+cot^2A}\) is equal to 

A. sec² A 

B. −1 

C. cot² A 

D. tan² A

Answer 1

To find: \(\frac{1+tan^2A}{1+cot^2A}\) 

Consider \(\frac{1+tan^2A}{1+cot^2A}\)

∵ 1 + tan²A = sec²A and 1 + cot²A = cosec²A

∴ \(\frac{1+tan^2A}{1+cot^2A}\) = \(\frac{sec^2A}{cosec^2A}\) = \(\frac{1/cos^2A}{1/sin^2A}\) \(\Big[∵secA\frac{1}{cosA}\,and\,cosecA=\frac{1}{sinA}\Big]\) 

= \(\frac{sin^2A}{cos^2A}\) = tan²A \(\Big[∵tanA = \frac{sinA}{cosA}\Big]\)

Answer 2

\(\frac{1+Tan^2θ}{1+Cot^2θ}\)

= \(\frac{Sec^2θ}{Cosec^2θ}\)

= \(\frac{Sin^2θ}{Cos^2θ} \)

= \(Tan^2θ\) (D)

Hope it helps

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