The value of tan⁡2tan−1⁡35+sin−1⁡513 is equal to

Let A=tan⁡2tan−1⁡35+sin−1⁡513 ...(i)Now, using 2tan−1⁡x=tan−1⁡2x1−x22tan−1⁡35=tan−1⁡2351−352=tan−1⁡651−925=tan−1⁡3016=tan−1⁡158Let sin−1⁡513=θ, then sin⁡θ=513∴ tan⁡θ=sin⁡θ1−sin2⁡θ=5131−532=5131213⇒ θ=tan−1⁡512=sin−1⁡513From Eq. (i),A=tan⁡tan−1⁡158+tan−1⁡512=tan⁡tan−1⁡158+5121−158512=tan⁡tan−1⁡55242196=tan⁡tan−1⁡55×421=tan⁡tan−1⁡22021=22021

The value of tan⁡2tan−1⁡35+sin−1⁡513 is equal to

Let A=tan⁡2tan−1⁡35+sin−1⁡513 ...(i)Now, using 2tan−1⁡x=tan−1⁡2x1−x22tan−1⁡35=tan−1⁡2351−352=tan−1⁡651−925=tan−1⁡3016=tan−1⁡158Let sin−1⁡513=θ, then sin⁡θ=513∴ tan⁡θ=sin⁡θ1−sin2⁡θ=5131−532=5131213⇒ θ=tan−1⁡512=sin−1⁡513From Eq. (i),A=tan⁡tan−1⁡158+tan−1⁡512=tan⁡tan−1⁡158+5121−158512=tan⁡tan−1⁡55242196=tan⁡tan−1⁡55×421=tan⁡tan−1⁡22021=22021

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The value of $\tan (2 \tan^{- 1} (\frac{3}{5}) + \sin^{- 1} (\frac{5}{13}))$ is equal to

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$\frac{- 181}{69}$

$\frac{151}{63}$

$\frac{- 291}{76}$

$\frac{220}{21}$

answer is B.

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Detailed Solution

Let $A = \tan (2 \tan^{- 1} (\frac{3}{5}) + \sin^{- 1} (\frac{5}{13}))$ ...(i)

Now, using $2 \tan^{- 1} x = \tan^{- 1} \frac{2 x}{1 - x^{2}}$

$\begin{matrix} 2 \tan^{- 1} (\frac{3}{5}) = \tan^{- 1} (\frac{2 (\frac{3}{5})}{1 - {(\frac{3}{5})}^{2}}) = \tan^{- 1} (\frac{\frac{6}{5}}{1 - \frac{9}{25}}) \\ = \tan^{- 1} (\frac{30}{16}) = \tan^{- 1} (\frac{15}{8}) \end{matrix}$

Let $\sin^{- 1} (\frac{5}{13}) = θ, then \sin θ = \frac{5}{13}$

$\begin{array}{l} ∴ \tan θ = \frac{\sin θ}{\sqrt{1 - \sin^{2} θ}} = \frac{\frac{5}{13}}{\sqrt{1 - {(\frac{5}{3})}^{2}}} = \frac{\frac{5}{13}}{\frac{12}{13}} \\ \Rightarrow θ = \tan^{- 1} (\frac{5}{12}) = \sin^{- 1} (\frac{5}{13}) \end{array}$

From Eq. (i),

$\begin{matrix} A = \tan (\tan^{- 1} (\frac{15}{8}) + \tan^{- 1} (\frac{5}{12})) \\ = \tan [\tan^{- 1} (\frac{\frac{15}{8} + \frac{5}{12}}{1 - (\frac{15}{8}) (\frac{5}{12})}]] \\ \begin{matrix} = \tan [\tan^{- 1} (\frac{\frac{55}{24}}{\frac{21}{96}})] = \tan [\tan^{- 1} \frac{55 \times 4}{21}] \\ = \tan (\tan^{- 1} \frac{220}{21}) = \frac{220}{21} \end{matrix} \end{matrix}$