Question

1+tan^2 0 / 1+cot^2 0
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Answer 1

Answer:

answer for the given problem is given

Answer 2

$\dfrac{1+{\tan}^2\theta}{1+{\cot}^2 \theta} = \dfrac{{\sec}^2\theta}{{\csc}^2\theta}$

$\dashrightarrow\dfrac{{\sec}^2\theta}{{\csc}^2 \theta} = \dfrac{\dfrac{1}{{\cos}^2\theta}}{\dfrac{1}{{\sin}^2\theta}} \dashrightarrow \dfrac{1}{\cos^2\theta} \times \dfrac{\sin^2\theta}{1}$

$\dashrightarrow\dfrac{{\sin}^2\theta}{{\cos}^2\theta} = \bf tan^2\theta$

Identities used:-

$1+\tan^2\theta=\sec^2\theta$

$1+\cot^2\theta=\csc^2\theta$

$\csc\theta=\dfrac{1}{\sin \theta}$

$\cot\theta=\dfrac{1}{\sec \theta}$

Also,

$\csc \theta = \rm cosec \ \theta$