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Question

Mathematics Question on Inverse Trigonometric Functions

If cos135+cos11213=cos1k\cos^{-1} \frac{3}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} k, then the value of kk is

A

1665\frac{16}{65}

B

1265\frac{12}{65}

C

1165\frac{11}{65}

D

1965\frac{19}{65}

Answer

1665\frac{16}{65}

Explanation

Solution

We have,
cos135+cos11213=cos1k\cos^{-1} \frac{3}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1}\,k
\Rightarrow \cos^{-1}\left\\{\frac{3}{5} \cdot \frac{12}{13}-\sqrt{1-\left(\frac{3}{5}\right)^{2}\sqrt{1-\left(\frac{12}{13}\right)^{2}}}\right\\} = \cos^{-1}\,k
[\because \cos^{-1}\,x + \cos^{-1}\, y = \cos^{-1}\left\\{xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}\right\\}, if
1x,y1-1 \le x, y \le 1 and x+y0]x+y \ge 0]
\Rightarrow \cos^{-1}\left\\{\frac{36}{65}-\frac{4}{5} \times \frac{5}{13}\right\\} = \cos^{-1}\,k
\Rightarrow \cos^{-1}\left\\{\frac{36}{65}-\frac{20}{65}\right\\} = \cos^{-1}\,k
cos11665=cos1k\Rightarrow \cos^{-1} \frac{16}{65} = \cos^{-1}\,k
k=1665\Rightarrow k=\frac{16}{65}