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Exercise 4.5 · Q8

Q.Simplify: tan⁡−1xy−tan⁡−1x−yx+y\tan^{-1}\dfrac{x}y - \tan^{-1}\dfrac{x-y}{x+y}.

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We apply tan⁡−1p−tan⁡−1q=tan⁡−1p−q1+pq\tan^{-1}p-\tan^{-1}q=\tan^{-1}\dfrac{p-q}{1+pq} symbolically with p=xy, q=x−yx+yp=\dfrac xy,\ q=\dfrac{x-y}{x+y}, and the resulting fraction simplifies to exactly 11.

Step 1. Set p=xy, q=x−yx+yp=\dfrac xy,\ q=\dfrac{x-y}{x+y}, and compute p−qp-q. p−q=xy−x−yx+y=x(x+y)−y(x−y)y(x+y)=x2+xy−xy+y2y(x+y)=x2+y2y(x+y)p-q=\dfrac xy-\dfrac{x-y}{x+y}=\dfrac{x(x+y)-y(x-y)}{y(x+y)}=\dfrac{x^2+xy-xy+y^2}{y(x+y)}=\dfrac{x^2+y^2}{y(x+y)}.

Step 2. Compute 1+pq1+pq. pq=xy⋅x−yx+y=x(x−y)y(x+y)pq=\dfrac xy\cdot\dfrac{x-y}{x+y}=\dfrac{x(x-y)}{y(x+y)}, so 1+pq=y(x+y)+x(x−y)y(x+y)=xy+y2+x2−xyy(x+y)=x2+y2y(x+y)1+pq=\dfrac{y(x+y)+x(x-y)}{y(x+y)}=\dfrac{xy+y^2+x^2-xy}{y(x+y)}=\dfrac{x^2+y^2}{y(x+y)}.

Step 3. Divide. p−q1+pq=(x2+y2)/(y(x+y))(x2+y2)/(y(x+y))=1\dfrac{p-q}{1+pq}=\dfrac{(x^2+y^2)/(y(x+y))}{(x^2+y^2)/(y(x+y))}=1. …

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