Inverse Trigonometric Identities

Before reading this, make sure you are familiar with inverse trigonometric functions.

The following inverse trigonometric identities give an angle in different ratios. Before the more complicated identities come some seemingly obvious ones. Be observant of the conditions the identities call for.

$$\begin{array}{ l l l } \sin^{-1}(-x) &=-\sin^{-1}x,~&\lvert x\rvert\leq1 \\ \cos^{-1}(-x) &=\pi-\cos^{-1}x,~&\lvert x\rvert\leq1 \\ \tan^{-1}(-x) &=-\tan^{-1}x,~&x\in\mathbb{R} \\ \cot^{-1}(-x) &=\pi-\cot^{-1}x,~&x\in\mathbb{R} \\ \csc^{-1}x &=\sin^{-1}\left(\frac{1}{x}\right),~&\lvert x\rvert\geq1 \\ \sec^{-1}x &=\cos^{-1}\left(\frac{1}{x}\right),~&\lvert x\rvert\geq1 \\ \cot^{-1}x &=\tan^{-1}\left(\frac{1}{x}\right),~&x>0 \\ \cot^{-1}x &=\pi+\tan^{-1}\left(\frac{1}{x}\right),~&x<0 \\ \sin^{-1}x+\cos^{-1}x &=\frac{\pi}{2},~&\lvert x\rvert\leq1 \\ \csc^{-1}x+\sec^{-1}x &=\frac{\pi}{2},~&\lvert x\rvert\geq1 \end{array}$$

Now for the more complicated identities. These come handy very often, and can easily be derived using the basic trigonometric identities.

$$\begin{array}{ l l l } \sin^{-1}x&=\cos^{-1}\left(\sqrt{1-x^{2}}\right),~&x\geq0 \\ \cos^{-1}x&=\sin^{-1}\left(\sqrt{1-x^{2}}\right),~&x\geq0 \\ \cos^{-1}x&=\pi-\sin^{-1}\left(\sqrt{1-x^{2}}\right),~&x<0 \\ \tan^{-1}x+\tan^{-1}y&=\tan^{-1}\left(\frac{x+y}{1-xy}\right),~&xy<1 \\ \tan^{-1}x+\tan^{-1}y&=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right),~&xy>1 \\ \tan^{-1}x-\tan^{-1}y&=\tan^{-1}\left(\frac{x-y}{1+xy}\right) \\ \sin^{-1}x&=\tan^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right),~&x\in(0,1) \\ \cos^{-1}x&=\tan^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right),~&x\in(0,1) \\ \tan^{-1}x&=\sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right),~&x>0 \\ \tan^{-1}x&=\cos^{-1}\left(\frac{1}{\sqrt{x^{2}+1}}\right),~&x>0 \end{array}$$

Find the value of $x$ for which $\sin\big(\cot^{-1}(1+x)\big)=\cos\big(\tan^{-1}(x)\big).$

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We have

$$\begin{aligned} \cot^{-1}(1+x)&=\sin^{-1}\left(\frac{1}{\sqrt{1+(1+x)^{2}}}\right) \\ \tan^{-1}x&=\cos^{-1}\left(\frac{1}{\sqrt{x^{2}-1}}\right). \end{aligned}$$

Therefore, we have

$$\begin{aligned} \sin\big(\cot^{-1}(1+x)\big)&=\cos\big(\tan^{-1}(x)\big) \\ \sin\left(\sin^{-1}\bigg(\frac{1}{\sqrt{1+(1+x)^{2}}}\bigg)\right)&=\cos\left(\cos^{-1}\bigg(\frac{1}{\sqrt{x^{2}-1}}\bigg)\right) \\ \frac{1}{\sqrt{1+(1+x)^{2}}} &= \frac{1}{\sqrt{x^{2}+1}} \\ x^{2}+1&=(x+1)^{2}+1 \\ x^{2}+2x+1&=x^{2} \\ x&=-\frac{1}{2}.\ _\square \end{aligned}$$