Trig. Problem

**raj01b** · August 23rd 2009, 11:52 AM

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

**red_dog** · August 23rd 2009, 12:05 PM

It's not an equation, it's an identity.

The left side member:

$\frac{1+\tan^2A}{1+\cot^2A}=\frac{1+\tan^2A}{1+\fr ac{1}{\tan^2A}}=\frac{1+\tan^2A}{\frac{1+\tan^2A}{ \tan^2A}}=\tan^2A$

$\left(\frac{1+\tan A}{1+\cot A}\right)^2=\left(\frac{1+\tan A}{1+\frac{1}{\tan A}}\right)^2=$

$=\left(\frac{1+\tan A}{\frac{1+\tan A}{\tan A}}\right)^2=\tan^2A$

Then the identity holds.

**Soroban** · August 23rd 2009, 03:27 PM

Hello, raj01b!

Another approach . . .

Prove: . $\frac{1+\tan^2\!A}{1+\cot^2\!A} \;=\;\left(\frac{1+\tan A}{1+\cot A}\right)^2$

The left side is: . $\frac{1 + \tan^2\!A}{1 + \cos^2\!A} \;=\;\frac{\sec^2\!A}{\csc^2\!A} \;=\; \frac{\dfrac{1}{\cos^2\!A}}{\dfrac{1}{\sin^2\!A}} \;=\;\frac{\sin^2\!A}{\cos^2\!A} \;=\;\tan^2\!A$

Multiply by $\frac{(1+\tan A)^2}{(1 +\tan A)^2}\!:\quad\tan^2\!A\cdot\frac{(1+\tan A)^2}{(1+\tan A)^2} \;=\;\left[\frac{\tan A(1 + \tan A)}{1 + \tan A}\right]^2$

Divide top and bottom by $\tan A\!:\quad\left(\frac{1+\tan A}{\frac{1}{\tan A} + 1}\right)^2 \;=\;\left(\frac{1+\tan A}{1 + \cot A}\right)^2$

**raj01b** · August 25th 2009, 04:27 AM

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

**Grandad** · August 25th 2009, 04:59 AM

Hello raj01b

Originally Posted by raj01b

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

You've already seen how $\frac{1+\tan^2A}{1+\cot^2A}= \tan^2A$

Well if we multiply $\Big(\frac{1-\tan A}{1-\cot A}\Big)^2$ 'top-and-bottom' by $\tan^2A$ , we get:

$\Big(\frac{1-\tan A}{1-\cot A}\Big)^2=\Big(\frac{1-\tan A}{1-\cot A}\Big)^2\times \frac{\tan^2A}{\tan^2A}$

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

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

$= \tan^2A$ , since $(1-\tan A)^2 = (\tan A -1)^2$

So $\frac{1+\tan^2A}{1+\cot^2A}=\Big(\frac{1-\tan A}{1-\cot A}\Big)^2$

Grandad

**raj01b** · August 25th 2009, 05:41 AM

Thanks Soroban for the quick reply. I was actually confused that (1-tan A)^2 & (tan A-1)^2 are the same thing.

Thank you very much once again.

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