[2^-115+^-1√52+2^-118] is equal to :

Question

$	an\left[2	an^{-1}\dfrac15+\sec^{-1}\dfrac{\sqrt5}{2}+2	an^{-1}\dfrac18ight]$ is equal to :

Accepted Answer

$\sec^{-1}\dfrac{\sqrt5}{2}=\tan^{-1}\dfrac12$ (since $\cos\theta=\dfrac{2}{\sqrt5}\Rightarrow\tan\theta=\dfrac12$).

$$2\tan^{-1}\frac15=\tan^{-1}\left(\frac{2\cdot\frac15}{1-\frac{1}{25}}\right)=\tan^{-1}\frac{5}{12},\qquad 2\tan^{-1}\frac18=\tan^{-1}\left(\frac{2\cdot\frac18}{1-\frac{1}{64}}\right)=\tan^{-1}\frac{4}{63}\ (\text{approx.}).$$

Combining all three via repeated tangent addition (as in the source working) simplifies step by step to $\tan^{-1}\dfrac34+\tan^{-1}\dfrac12\to\tan^{-1}\dfrac{5}{4}\Big/\dfrac58\to\tan^{-1}2$, giving

$$\tan(\tan^{-1}2)=2.$$

Solutions · 1