Question

1+tan^2x/1+cot^2x=(1-tanx/1-cotx)^2 prove that

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

The given equation is: $\frac{1+tan^{2}x}{1+cot^{2}x}=(\frac{1-tanx}{1-cotx}) ^{2}$.

Taking the LHS of the given equation,

$\frac{1+tan^{2}x}{1+cot^{2}x}=\frac{sec^{2}x}{cosec^{2}x}=\frac{sin^{2}x}{cos^{2}x}=tan^{2}x$

Taking the RHS of the given equation,

$(\frac{1-tanx}{1-cotx}) ^{2}$=$(\frac{1-tanx}{1-\frac{1}{tanx}})^{2}$=$(\frac{(tanx)(1-tanx)}{1-tanx}) ^{2}$=$tan^{2}x$

Hence, LHS=RHS