Sum and difference of angles in terms of arctan - Practice Questions & MCQ

The sum of angles in terms of arctan
1. $\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\left\{\begin{array}{cc}\tan ^{-1}\left(\frac{x+y}{1-x y}\right), & \text { If } \mathrm{x}>0, y>0, x y<1 \\ \pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right), & \text { If } \mathrm{x}>0, \mathrm{y}>0 \text { and } \mathrm{xy}>1 \\ -\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right), & \text { If } \mathrm{x}<0, \mathrm{y}<0 \text { and } \mathrm{xy}>1\end{array}\right.$

Sum of angles in terms of arctan
1. $\tan ^{-1} \mathrm{x}-\tan ^{-1} \mathrm{y}=\left\{\begin{array}{cc}\tan ^{-1}\left(\frac{x-y}{1+x y}\right), & \text { If } x y>-1 \\ \pi+\tan ^{-1}\left(\frac{x-y}{1+x y}\right), & \text { If } \mathrm{x}>0, \mathrm{y}<0 \text { and } \mathrm{xy}<-1 \\ -\pi+\tan ^{-1}\left(\frac{x-y}{1+x y}\right), & \text { If } \mathrm{x}<0, \mathrm{y}>0 \text { and } \mathrm{xy}<-1\end{array}\right.$

Sum and difference of angles in terms of arcsin
1. $\sin ^{-1} \mathrm{x}+\sin ^{-1} \mathrm{y}= \begin{cases} & \text { if }-1 \leq x, y \leq 1 \text { and } x^2+y^2 \leq 1 \\ \sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\} & \text { or, if } x y<0 \text { and } x^2+y^2>1 \\ \pi-\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\} & \text { if } 0<x, y \leq 1 \text { and } x^2+y^2>1 \\ -\pi-\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\} & \text { if }-1 \leq x, y<0 \text { and } x^2+y^2>1\end{cases}$
2. $\sin ^{-1} \mathrm{x}-\sin ^{-1} \mathrm{y}= \begin{cases} & \text { if }-1 \leq x, y \leq 1 \text { and } x^2+y^2 \leq 1 \\ \sin ^{-1}\left\{x \sqrt{1-y^2}-y \sqrt{1-x^2}\right\} & \text { or, if } x y>0 \text { and } x^2+y^2>1 \\ \pi-\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\} & \text { if } 0<x \leq 1 ;-1 \leq y<0 \text { and } x^2+y^2>1 \\ -\pi-\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\} & \text { if }-1 \leq x<0 ; 0<y \leq 1 \text { and } x^2+y^2>1\end{cases}$

Sum and difference of angles in terms of arccos
1. $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left\{x y-\sqrt{1-x^2} \sqrt{1-y^2}\right\} \quad$ if $0<x, y \leq 1$
2. $\cos ^{-1} x-\cos ^{-1} y= \begin{cases}\cos ^{-1}\left\{x y+\sqrt{1-x^2} \sqrt{1-y^2}\right\} & \text { if } 0 \leq x, y \leq 1 \text { and } x \leq y \\ -\cos ^{-1}\left\{x y+\sqrt{1-x^2} \sqrt{1-y^2}\right\} & \text { if } 0<x, y \leq 1 \text { and } x>y\end{cases}$