If the inverse trigonometric functions take principal values, then cos-1310costan-143+25sintan-143 is equal to :

Lettan-143=θ⇒tanθ=43E=cos-1310cosθ+25sinθ =cos-1310×35+25·45 =cos-1950+825=cos-12550=cos-112=π3

If the inverse trigonometric functions take principal values, then cos-1310costan-143+25sintan-143 is equal to :

Lettan-143=θ⇒tanθ=43E=cos-1310cosθ+25sinθ =cos-1310×35+25·45 =cos-1950+825=cos-12550=cos-112=π3

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If the inverse trigonometric functions take principal values, then $\cos^{- 1} (\frac{3}{10} \cos (\tan^{- 1} (\frac{4}{3})) + \frac{2}{5} \sin (\tan^{- 1} (\frac{4}{3})))$ is equal to :

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$\frac{π}{4}$

$\frac{π}{6}$

$\frac{π}{3}$

answer is C.

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Detailed Solution

Let

$\tan^{- 1} \frac{4}{3} = θ \Rightarrow \tan θ = \frac{4}{3}$

$E = \cos^{- 1} (\frac{3}{10} \cos θ + \frac{2}{5} \sin θ) = \cos^{- 1} (\frac{3}{10} \times \frac{3}{5} + \frac{2}{5} \cdot \frac{4}{5}) = \cos^{- 1} (\frac{9}{50} + \frac{8}{25}) = \cos^{- 1} (\frac{25}{50}) = \cos^{- 1} (\frac{1}{2}) = \frac{π}{3}$