NDA 2012
Exam NDA
Year 2012
Subject Mathematics
Chapter Inverse Trigonometric Functions NDA 2012 · Mathematics · Inverse Trigonometric Functions
What is the value of \ ^ -1 (4)/(5)+ ^ -1 (12)/(13)\ ?
What is the value of
cos { cos − 1 4 5 + cos − 1 12 13 } ? \cos \left\{\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}\right\} ? cos { cos − 1 5 4 + cos − 1 13 12 } ? Options A 63/65 B 33/65 C 22/65 D 11/65
Step-by-step solution
Put cos − 1 4 5 = A cos A = 4 5 ⇒ sin A = 3 5 cos − 1 12 13 = B cos B = 12 13 ⇒ sin B = 5 13 . Now, cos ( cos − 1 4 5 + cos − 1 12 13 ) = cos ( A + B ) = cos A cos B − sin A sin B = ( 4 5 ) ( 12 13 ) − ( 3 5 ) ( 5 13 ) = 48 65 − 15 65 = 33 65 . \begin{aligned}
& \text{ Put } \cos ^{-1} \frac{4}{5}=\mathrm{A} \\
& \cos \mathrm{ A}=\frac{4}{5} \\
& \Rightarrow \quad \sin \mathrm{ A}=\frac{3}{5} \\
& \cos ^{-1} \frac{12}{13}=\mathrm{B} \\
& \cos \mathrm{ B}=\frac{12}{13} \\
& \Rightarrow \quad \sin \mathrm{ B}=\frac{5}{13} \text{. } \\
& \text{ Now, } \\
& \cos \left(\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}\right) \\
& =\cos (\mathrm{A}+\mathrm{B}) \\
& =\cos \mathrm{A} \cos \mathrm{ B}-\sin \mathrm{A} \sin \mathrm{ B} \\
& =\left(\frac{4}{5}\right)\left(\frac{12}{13}\right)-\left(\frac{3}{5}\right)\left(\frac{5}{13}\right) \\
& =\frac{48}{65}-\frac{15}{65}=\frac{33}{65}.
\end{aligned} Put cos − 1 5 4 = A cos A = 5 4 ⇒ sin A = 5 3 cos − 1 13 12 = B cos B = 13 12 ⇒ sin B = 13 5 . Now, cos ( cos − 1 5 4 + cos − 1 13 12 ) = cos ( A + B ) = cos A cos B − sin A sin B = ( 5 4 ) ( 13 12 ) − ( 5 3 ) ( 13 5 ) = 65 48 − 65 15 = 65 33 .
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