MHT CET · Maths · Inverse Trigonometric Functions
\( \cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)= \)
- A \(\cos ^{-1}\left(\frac{24}{25}\right)\)
- B \(\cos ^{-1}\left(\frac{33}{65}\right)\)
- C \(\cos ^{-1}\left(\frac{5}{13}\right)\)
- D \(\cos ^{-1}\left(\frac{3}{5}\right)\)
Answer & Solution
Correct Answer
(B) \(\cos ^{-1}\left(\frac{33}{65}\right)\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right) \\
& =\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right) \\
& =\tan ^{-1}\left[\frac{\left(\frac{3}{4}\right)+\left(\frac{5}{12}\right)}{1-\left(\frac{3}{12}\right)\left(\frac{5}{12}\right)}\right] \\
& =\tan ^{-1}\left(\frac{36+20}{48-15}\right)=\tan ^{-1}\left(\frac{56}{13}\right)=\cos ^{-1}\left(\frac{33}{65}\right)
\end{aligned}
\)
\begin{aligned}
& \cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right) \\
& =\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right) \\
& =\tan ^{-1}\left[\frac{\left(\frac{3}{4}\right)+\left(\frac{5}{12}\right)}{1-\left(\frac{3}{12}\right)\left(\frac{5}{12}\right)}\right] \\
& =\tan ^{-1}\left(\frac{36+20}{48-15}\right)=\tan ^{-1}\left(\frac{56}{13}\right)=\cos ^{-1}\left(\frac{33}{65}\right)
\end{aligned}
\)
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