Inverse Circular Functions 3 Question 9
9. The value of $\tan \cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}$ is
(1983, 1M)
(a) $\frac{6}{17}$
(b) $\frac{17}{6}$
(c) $\frac{16}{7}$
(d) None of these
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Answer:
Correct Answer: 9. (b)
Solution:
- $\tan \cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}=\tan \tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}$
$$ \because \cos ^{-1} \frac{4}{5}=\tan ^{-1} \frac{3}{4} $$
$$ =\tan \tan ^{-1} \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}=\tan \tan ^{-1} \frac{17}{6}=\frac{17}{6} $$
