JEE Advance - Mathematics (2024 - Paper 2 Online - No. 1)
Considering only the principal values of the inverse trigonometric functions, the value of
$$ \tan \left(\sin ^{-1}\left(\frac{3}{5}\right)-2 \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right) $$
is
$\frac{7}{24}$
$\frac{-7}{24}$
$\frac{-5}{24}$
$\frac{5}{24}$
Explanation
$$\begin{aligned}
& \tan \left(\tan ^{-1}\left(\frac{3}{4}\right)-2 \tan ^{-1}\left(\frac{1}{2}\right)\right) \\
& \tan \left(\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{2 \times \frac{1}{2}}{1-\frac{1}{4}}\right)\right) \\
& \tan \left(\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{4}{3}\right)\right) \\
& \tan \left(\tan ^{-1}\left(\frac{3}{4}-\frac{4}{3}\right)\right) \\
& \frac{9-16}{24}=\frac{-7}{24}
\end{aligned}$$
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