Question

tan-1 63/16 = sin-1 5/13 +cos-1 ⅗

Answer 1

Answer:

Step-by-step explanation:

माना    $sin^{-1}(\dfrac{5}{13} )=\theta, cos^{-1}(\dfrac{3}{5} )=Q$

तब   $tan\theta=\dfrac{sin\theta}{cos\theta} = \dfrac{sin(sin^{-1}(\dfrac{5}{13} ))}{cos(sin^{-1}(\dfrac{5}{13})) }\\\\\\=\dfrac{\dfrac{5}{13} }{\sqrt{1-(5/13)^2} } \\\\\\=\dfrac{5/13}{12/13} \\\\\\=\dfrac{5}{12}$

और   $tanQ=\dfrac{sinQ}{cosQ} = \dfrac{sin(cos^{-1}(\dfrac{3}{5} ))}{cos(cos^{-1}(\dfrac{3}{5})) }\\\\\\=\dfrac{\sqrt{1-(3/5)^2} }{3/5} \\\\\\=\dfrac{4/5}{3/5} \\\\\\=\dfrac{4}{3}$

इसलिए

        $tan(\theta+Q)=tan\dfrac{tan\theta+tanQ}{1-tan\theta tanQ} \\\\\\=\dfrac{\dfrac{4}{12}+\dfrac{4}{3}  }{1-\dfrac{5}{12} *\dfrac{4}{5} } \\\\\\=\dfrac{15+48}{36-20} \\\\=\dfrac{63}{16} \\\\\\=>\theta+Q=tan^{-1}(\dfrac{63}{16} )$

Answer 2

$let \: \: \: \: {sin}^{ - 1} \frac{5}{13} = a \: \: \: \: \: \: {cos}^{ - 1} \frac{3}{5} = b \\ \\ sina = \frac{5}{13} \\ \\ cosa = \sqrt{1 - \frac{25}{169} } \\ \\ cosa = \sqrt{ \frac{144}{169} } \\ \\ cosa = \frac{12}{13} \\ \\ {cos}^{ - 1} \frac{3}{5} = b \\ \\ cosb = \frac{3}{5} \\ \\ sinb = \sqrt{1 - \frac{9}{25} } \\ \\ sinb = \frac{4}{5} \\ \\ tana = \frac{sina}{cosa} = \frac{ \frac{5}{13} }{ \frac{12}{13} } = \frac{5}{12} \\ \\ tanb = \frac{ \frac{4}{5} }{ \frac{3}{5} } = \frac{4}{3} \\ \\ tan(a + b) = \frac{tana + tanb}{1 - tanatanb} \\ \\ tan(a + b) = \frac{ \frac{5}{12} + \frac{4}{3} }{1 - \frac{5}{12} \times \frac{4}{3} } \\ \\ tan(a + b) = \frac{15 + 48}{36 - 20} \\ \\ {tan}(a + b) = \frac{63}{16} \\ \\ a + b = {tan}^{ - 1} ( \frac{63}{16}) \\ \\ {sin}^{ - 1} \frac{5}{13} + {cos}^{ - 1} \frac{3}{5} = {tan}^{ - 1} \frac{63}{16}$