Evaluating Composite Inverse Expressions

• Nested evaluation: name the inner inverse $\theta = \csc^{-1}2$ (so $\theta = \tfrac{\pi}{6}$), then work outward: $\cot\theta$, then $\tan^{-1}(\cot\theta)$.
• $\sin^{-1}(\sin x)$: equals $x$ only for $x \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$. Otherwise use $\sin x = \sin(\pi - x)$ to bring the angle into range (e.g. $\sin^{-1}(\sin\tfrac{2\pi}{3}) = \tfrac{\pi}{3}$).
• For $\cot^2(\sec^{-1}2) + \tan^2(\csc^{-1}\sqrt3)$-type sums, evaluate each inverse to a standard angle first.

The ordinary trigonometry that fires the moment $\theta$ is named (the step after "name the inner angle" in three PYQs out of every ten here): $\sin^2\theta+\cos^2\theta=1$ to get the OTHER ratio from the one the inverse hands you ($\theta=\sin^{-1}\tfrac35\Rightarrow\cos\theta=\tfrac45$, positive because $\theta\in[-\tfrac\pi2,\tfrac\pi2]$); $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$ and $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ for $\sin(\sin^{-1}\tfrac35+\cos^{-1}\tfrac{12}{13})$-type sums; $\sin3\theta=3\sin\theta-4\sin^3\theta$ and $\cos3\theta=4\cos^3\theta-3\cos\theta$ for $\sin(3\sin^{-1}x)$; and the cotangent forms $\cot2\theta=\dfrac{\cot^2\theta-1}{2\cot\theta}$, $\cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$ when a question is posed entirely in $\cot^{-1}$.

Set $\theta = (\text{the inner inverse})$, so its argument gives $\tan\theta$ (or $\sin\theta, \cos\theta$ via a triangle), then apply:

• Double angle: $\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}$, $\sin 2\theta = 2\sin\theta\cos\theta$.
• Half angle: $\tan\tfrac{\theta}{2} = \dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1+\cos\theta}$, and squared, $\tan^2\tfrac\theta2=\dfrac{1-\cos\theta}{1+\cos\theta}$ — the form to use when the inner inverse gives $\cos\theta$ directly ($\theta=\cos^{-1}x\Rightarrow\tan^2\tfrac\theta2=\dfrac{1-x}{1+x}$). Also $\cos^2\tfrac\theta2=\dfrac{1+\cos\theta}{2}$, $\sin^2\tfrac\theta2=\dfrac{1-\cos\theta}{2}$.

Useful for $\tan(2\tan^{-1}x)$, $\tan\!\left(\tfrac12 \sec^{-1}t\right)$, and $\sqrt{1 + \sin(2\cos^{-1}t)}$.

For each inverse, build the right triangle to read its tangent: e.g. $\sin^{-1}\tfrac35 \Rightarrow \tan = \tfrac34$; $\cot^{-1}\tfrac32 \Rightarrow \tan = \tfrac23$; $\csc^{-1}\tfrac{\sqrt{41}}{4} \Rightarrow \tan = \tfrac45$. Then combine the resulting $\tan^{-1}$ terms with the sum/difference formula, and apply the outer function ($\cot$, $\tan$, etc.). Or convert directly by reciprocals — often shorter than the triangle: $\sec^{-1}x=\cos^{-1}\tfrac1x$, $\csc^{-1}x=\sin^{-1}\tfrac1x$ (all $|x|\ge1$), $\cot^{-1}x=\tan^{-1}\tfrac1x$ (for $x>0$). So $\sec^{-1}\tfrac54=\cos^{-1}\tfrac45$, and $\sec^{-1}\tfrac54+\sin^{-1}\tfrac45$ is a complementary pair $=\tfrac\pi2$ at sight; $\csc^{-1}2=\sin^{-1}\tfrac12=\tfrac\pi6$.