Question

prove that 1+ tan² theta upon 1+cot² theta =(1-tan theta upon 1-cot theta )²

Answer 1

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here is your answer

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Answer 2

Answer:

Step-by-step explanation:

The given equation is:

$\frac{1+tan^2{\theta}}{1+cot^2{theta}}=(\frac{1-tan{\theta}}{1-cot{\theta}})^{2}$

Taking the LHS of the above equation, we have

$\frac{1+tan^2{\theta}}{1+cot^2{theta}}$

=$\frac{sec^2\theta}{cosec^2\theta}$

=$\frac{sin^2\theta}{cos^2\theta}$

=$tan^2{\theta}$

Now, taking the RHS of the above equation, we have

$(\frac{1-tan{\theta}}{1-cot{\theta}})^{2}$

=$(\frac{1-tan{\theta}}{1-\frac{1}{tan\theta}})^{2}$

=$(\frac{(1-tan{\theta})tan{\theta}}{-(1-tan{\theta})})^2$

=$tan^2{\theta}$

Hence, LHS=RHS, thus proved.