The value of cos⁡cos−1⁡1215+cos−1⁡45 is

Let θ1=cos−1⁡1215∴cos⁡θ1=1215 and θ2=cos−1⁡45 ∴cos⁡θ2=45∴sin⁡θ1=915 and sin⁡θ2=35, assuming that θ1and θ2 are in the I quadrant∴cos⁡θ1+θ2=cos⁡θ1cos⁡θ2−sin⁡θ1sin⁡θ2 =1215×45−915×35=725.

The value of cos⁡cos−1⁡1215+cos−1⁡45 is

Let θ1=cos−1⁡1215∴cos⁡θ1=1215 and θ2=cos−1⁡45 ∴cos⁡θ2=45∴sin⁡θ1=915 and sin⁡θ2=35, assuming that θ1and θ2 are in the I quadrant∴cos⁡θ1+θ2=cos⁡θ1cos⁡θ2−sin⁡θ1sin⁡θ2 =1215×45−915×35=725.

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The value of $\cos (\cos^{- 1} (\frac{12}{15}) + \cos^{- 1} (\frac{4}{5}))$ is

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$\frac{1}{5}$

$\frac{3}{10}$

$\frac{3}{25}$

$\frac{7}{25}$

answer is D.

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Detailed Solution

Let $θ_{1} = \cos^{- 1} \frac{12}{15} ∴ \cos θ_{1} = \frac{12}{15}$

$and θ_{2} = \cos^{- 1} \frac{4}{5} ∴ \cos θ_{2} = \frac{4}{5}$

$∴ \sin θ_{1} = \frac{9}{15} and \sin θ_{2} = \frac{3}{5},$ assuming that $θ_{1}$

and $θ_{2}$ are in the I quadrant

$\begin{array}{c} ∴ \cos (θ_{1} + θ_{2}) = \cos θ_{1} \cos θ_{2} - \sin θ_{1} \sin θ_{2} \\ = \frac{12}{15} \times \frac{4}{5} - \frac{9}{15} \times \frac{3}{5} = \frac{7}{25} . \end{array}$