Please help solving (prove) following equation-
1+tan^2A/1+cot^2A=(1-tanA/1-cotA)^2
please read above as tan square A & cot square A
Thanks
It's not an equation, it's an identity.
The left side member:
$\displaystyle \frac{1+\tan^2A}{1+\cot^2A}=\frac{1+\tan^2A}{1+\fr ac{1}{\tan^2A}}=\frac{1+\tan^2A}{\frac{1+\tan^2A}{ \tan^2A}}=\tan^2A$
$\displaystyle \left(\frac{1+\tan A}{1+\cot A}\right)^2=\left(\frac{1+\tan A}{1+\frac{1}{\tan A}}\right)^2=$
$\displaystyle =\left(\frac{1+\tan A}{\frac{1+\tan A}{\tan A}}\right)^2=\tan^2A$
Then the identity holds.
Hello, raj01b!
Another approach . . .
The left side is: .$\displaystyle \frac{1 + \tan^2\!A}{1 + \cos^2\!A} \;=\;\frac{\sec^2\!A}{\csc^2\!A} \;=\; \frac{\dfrac{1}{\cos^2\!A}}{\dfrac{1}{\sin^2\!A}} \;=\;\frac{\sin^2\!A}{\cos^2\!A} \;=\;\tan^2\!A$Prove: .$\displaystyle \frac{1+\tan^2\!A}{1+\cot^2\!A} \;=\;\left(\frac{1+\tan A}{1+\cot A}\right)^2 $
Multiply by $\displaystyle \frac{(1+\tan A)^2}{(1 +\tan A)^2}\!:\quad\tan^2\!A\cdot\frac{(1+\tan A)^2}{(1+\tan A)^2} \;=\;\left[\frac{\tan A(1 + \tan A)}{1 + \tan A}\right]^2$
Divide top and bottom by $\displaystyle \tan A\!:\quad\left(\frac{1+\tan A}{\frac{1}{\tan A} + 1}\right)^2 \;=\;\left(\frac{1+\tan A}{1 + \cot A}\right)^2$
Thanks for the help but equation I posted earlier was not the equation I was stuck with, actually following is the equation I want to solve-
1+tan^2 A/1+cot^2 A= (1-tan A/1-cot A)^2
sorry for the trouble, any further help will be appreciated.
Thanks
Hello raj01bYou've already seen how $\displaystyle \frac{1+\tan^2A}{1+\cot^2A}= \tan^2A$
Well if we multiply $\displaystyle \Big(\frac{1-\tan A}{1-\cot A}\Big)^2$ 'top-and-bottom' by $\displaystyle \tan^2A$, we get:
$\displaystyle \Big(\frac{1-\tan A}{1-\cot A}\Big)^2=\Big(\frac{1-\tan A}{1-\cot A}\Big)^2\times \frac{\tan^2A}{\tan^2A}$
$\displaystyle = \frac{\tan^2A(1-\tan A)^2}{(\tan A - \cot A\tan A)^2}$
$\displaystyle = \frac{\tan^2A(1-\tan A)^2}{(\tan A - 1)^2}$
$\displaystyle = \tan^2A$, since $\displaystyle (1-\tan A)^2 = (\tan A -1)^2$
So $\displaystyle \frac{1+\tan^2A}{1+\cot^2A}=\Big(\frac{1-\tan A}{1-\cot A}\Big)^2$
Grandad