Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I solved a partial derivative problem and have the correct answer but the book I am using, Courant's Differential and Integral Calculus, has the answer in algebraic rather than trigonometric form. I never really worked on learning the translation of trig equations to algebraic form (except in the most basic cases). Anyway, my answer is in the form:

$$\frac{\sec^2(\arctan(x)+\arctan(y))}{(x^2+1)}$$

Courant has:

$$\frac{1+y^2}{(1-xy)^2}$$

I'd appreciate if someone could show me the translation presumably using right triangles and also perhaps suggest a good book or resource to work out similar problems. I need to bone up on this aspect of trig.

Thanks in advance

share|improve this question

1 Answer 1

up vote 3 down vote accepted

$\displaystyle\frac{\sec^2(\arctan(x)+\arctan(y))}{(x^2+1)}=\frac{\sec^2(\arctan(\frac{x+y}{1-xy}))}{(x^2+1)}=\frac{1+\tan^2(\arctan(\frac{x+y}{1-xy}))}{1+x^2}=\frac{1+(\frac{x+y}{1-xy})^2}{1+x^2}=\frac{y^2+1}{(1-xy)^2}$

share|improve this answer
    
Thanks Abhra, very nice, appreciate it. There are quite a few things going on in this answer that I wish I knew the identities for. Even the first step in your transformation I am embarrassed to say I don't know. The remaining steps I fully understand. Any suggestions on a good resource that I could study? –  Joe Feb 10 '13 at 7:37
    
I guess I should rephrase my comment: What is the identity or property of arctan(x) + arctan(y) that allows it to be transformed to arctan((x+y)/(1-xy))? I don't know that one. –  Joe Feb 10 '13 at 7:50
    
Well, I see from Wikipedia it is a standard arctan addition identity I didn't know. Shame on me. I really need to memorize more identities. Thanks again. –  Joe Feb 10 '13 at 8:01
    
Joe, it's the arctan version of the identity $\tan(a+b)=(\tan a+\tan b)/(1-\tan a\tan b)$. –  Gerry Myerson Feb 10 '13 at 9:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.