If α=cos−1⁡35, β=tan−1⁡13, where 0<α, β<π2 then α−β is equal to

α-β=cos-135-tan-113 ⇒α-β=tan-143-tan-113 ⇒α-β=tan-143-131+4313=tan-14-139+4=tan-1913=sin-19510

If α=cos−1⁡35, β=tan−1⁡13, where 0<α, β<π2 then α−β is equal to

α-β=cos-135-tan-113 ⇒α-β=tan-143-tan-113 ⇒α-β=tan-143-131+4313=tan-14-139+4=tan-1913=sin-19510

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If $α = \cos^{- 1} (\frac{3}{5}), β = \tan^{- 1} (\frac{1}{3})$ , where $0 < α, β < \frac{π}{2}$ then $α - β$ is equal to

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$\tan^{- 1} (\frac{9}{14})$

$\sin^{- 1} (\frac{9}{5 \sqrt{10}})$

$\cos^{- 1} (\frac{9}{5 \sqrt{10}})$

$\tan^{- 1} (\frac{9}{5 \sqrt{10}})$

answer is D.

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Detailed Solution

$α - β = \cos^{- 1} (\frac{3}{5}) - \tan^{- 1} (\frac{1}{3}) \Rightarrow α - β = \tan^{- 1} (\frac{4}{3}) - \tan^{- 1} (\frac{1}{3}) \Rightarrow α - β = \tan^{- 1} (\frac{(\frac{4}{3}) - (\frac{1}{3})}{1 + (\frac{4}{3}) (\frac{1}{3})}) = \tan^{- 1} \frac{(4 - 1) 3}{9 + 4} = \tan^{- 1} (\frac{9}{13}) = \sin^{- 1} (\frac{9}{5 \sqrt{10}})$