Question
Prove that $\left(\frac{1+\tan ^2 A}{1+\cot ^2 A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^2=\tan ^2 A$
✓
Answer
First, we will show that, $\frac{1+\tan ^2 A}{1+\cot ^2 A}=\tan ^2 A$
$\text { LHS }=\frac{1+\tan ^2 A}{1+\cot ^2 A}$
$=\frac{1+\tan ^2 A}{1+\frac{1}{\tan ^2 A}}$
$=\frac{1+\tan ^2 A}{\frac{1+\tan ^2 A}{\tan ^2 A}}$
$=\left(1+\tan ^2 A\right) \times \frac{\tan ^2 A}{1+\tan ^2 A}$
$=\tan ^2 A=\text { RHS } \ldots . \text { (i) }$
Now, we will show that, $\left(\frac{1-\tan A}{1-\cot A}\right)^2=\tan ^2 A$
$\text { LHS }=\left(\frac{1-\tan A}{1-\cot A}\right)^2$
$=\left(\frac{1-\tan A}{1-\frac{1}{\tan A}}\right)^2$
$=\left(\frac{1-\tan A}{\frac{\tan A-1}{\tan A}}\right)^2$
$=\left[(1-\tan A) \times\left(\frac{\tan A}{-(1-\tan A)}\right)\right]^2$
$=(\tan A)^2$
$=\tan ^2 A$
$=\text { RHS } \ldots \text { (ii) }$
Hence, from $(i)$ and $(ii),$
$\frac{1+\tan ^2 A}{1+\cot ^2 A}$
$=\left(\frac{1-\tan A}{1-\cot A}\right)^2$
$=\tan ^2 A$
Hence proved.
$\text { LHS }=\frac{1+\tan ^2 A}{1+\cot ^2 A}$
$=\frac{1+\tan ^2 A}{1+\frac{1}{\tan ^2 A}}$
$=\frac{1+\tan ^2 A}{\frac{1+\tan ^2 A}{\tan ^2 A}}$
$=\left(1+\tan ^2 A\right) \times \frac{\tan ^2 A}{1+\tan ^2 A}$
$=\tan ^2 A=\text { RHS } \ldots . \text { (i) }$
Now, we will show that, $\left(\frac{1-\tan A}{1-\cot A}\right)^2=\tan ^2 A$
$\text { LHS }=\left(\frac{1-\tan A}{1-\cot A}\right)^2$
$=\left(\frac{1-\tan A}{1-\frac{1}{\tan A}}\right)^2$
$=\left(\frac{1-\tan A}{\frac{\tan A-1}{\tan A}}\right)^2$
$=\left[(1-\tan A) \times\left(\frac{\tan A}{-(1-\tan A)}\right)\right]^2$
$=(\tan A)^2$
$=\tan ^2 A$
$=\text { RHS } \ldots \text { (ii) }$
Hence, from $(i)$ and $(ii),$
$\frac{1+\tan ^2 A}{1+\cot ^2 A}$
$=\left(\frac{1-\tan A}{1-\cot A}\right)^2$
$=\tan ^2 A$
Hence proved.
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