VIT - VITEEE 2025
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Multiple angles in terms of arctan and arccos is considered one the most difficult concept.
24 Questions around this concept.
Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying $\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}$ is :
If $|\mathrm{x}| \leq 1$, then $2 \tan ^2 x+\sin ^{-1} \frac{2 x}{1+x^2}$ is equal to
Multiple angles in terms of arcsin
1. $2 \sin ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), & \frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \\ \pi-\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), & x>\frac{1}{\sqrt{2}} \\ -\pi-\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), & x<-\frac{1}{\sqrt{2}}\end{array}\right.$
2. $3 \sin ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\sin ^{-1}\left(3 x-4 x^3\right), & -\frac{1}{2} \leq x \leq \frac{1}{2} \\ \pi-\sin ^{-1}\left(3 x-4 x^3\right), & x>\frac{1}{2} \\ -\pi-\sin ^{-1}\left(3 x-4 x^3\right) & x:-\frac{1}{2}\end{array}\right.$
Multiple angles in terms of arccos
1. $2 \cos ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\cos ^{-1}\left(2 x^2-1\right), & \text { if } 0 \leq x \leq 1 \\ 2 \pi-\cos ^{-1}\left(2 x^2-1\right), & \text { if }-1 \leq x \leq 0\end{array}\right.$
2. $3 \cos ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\cos ^{-1}\left(4 x^3-3 x\right), & \text { if } \frac{1}{2} \leq x \leq 1 \\ 2 \pi-\cos ^{-1}\left(4 x^3-3 x\right), & \text { if }-\frac{1}{2} \leq x \leq \frac{1}{2} \\ 2 \pi+\cos ^{-1}\left(4 x^3-3 x\right), & \text { if }-1 \leq x \leq-\frac{1}{2}\end{array}\right.$
Multiple angles in terms of arctan and arcsin
$
2 \tan ^{-1} \mathrm{x}=\left\{\begin{array}{cc}
\sin ^{-1}\left(\frac{2 x}{1+x^2}\right), & \text { if }-1 \leq x \leq 1 \\
\pi-\sin ^{-1}\left(\frac{2 x}{1+x^2}\right), & \text { if } x>1 \\
-\pi-\sin ^{-1}\left(\frac{2 x}{1+x^2}\right), & \text { if } x<-1
\end{array}\right.
$
Multiple angles in terms of arctan and arccos
$
2 \tan ^{-1} \mathrm{x}=\left\{\begin{array}{cl}
\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right), & \text { if } 0 \leq x<\infty \\
-\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right), & \text { if }-\infty<x \leq 0
\end{array}\right.
$
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