Question

Cos-1 12/13 + sin-1 ⅗ = sin-1 56/65

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

माना   $L.H.S.=\theta$  तब   $\theta>0$

और  

         $\theta=cos^{-1}(\dfrac{12}{13} )+sin^{-1}(\dfrac{3}{5} )\\\\\\<cos^{-1}\dfrac{1}{\sqrt{2} } +sin^{-1}\dfrac{1}{\sqrt{2} }=\dfrac{\pi}{4} +\dfrac{\pi}{4} \\\\\\(\dfrac{12}{13} >\dfrac{1}{\sqrt{2} } ;\dfrac{3}{5} <\dfrac{1}{\sqrt{2} } )\\\\\\sin\theta=sin(cos^{-1}(\dfrac{12}{13} )+sin^{-1}(\dfrac{3}{5} ))\\\\\\=sin(cos^{-1}(\dfrac{12}{13} )cos(sin^{-1}(\dfrac{3}{5} ))+cos(cos^{-1}(\dfrac{12}{13} ))sin(sin^{-1}(\dfrac{3}{5} ))$

$=\sqrt{1-(12/13)^2} \sqrt{1-(3/5)^2} +\dfrac{12}{13} *\dfrac{3}{5} \\\\\\=\dfrac{5}{13} *\dfrac{4}{5} +\dfrac{36}{65} \\\\\\=\dfrac{20+36}{65} \\\\\\=\dfrac{56}{65} \\\\\\sin\theta=\dfrac{56}{65}=>\theta=sin^{-1}(\dfrac{56}{65})$