Question

9π/8 - 9/4sin-1 ⅓ = 9/4 sin-1 2√2/3

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

$\dfrac{9\pi}{8} -\dfrac{9}{4} sin^{-1}(\dfrac{1}{3} )=\dfrac{9}{4} sin^{-1}(\dfrac{2\sqrt{2} }{3} )$

$L.H.S.=\dfrac{9\pi}{8} -\dfrac{9}{4} sin^{-1}(\dfrac{1}{3} )\\\\\\=\dfrac{9}{4} \{\dfrac{\pi}{2} -sin^{-1}\dfrac{1}{3} \}\\\\\\=\dfrac{9}{4}\{cos^{-1}(\dfrac{1}{3} )\}\\\\\\(sin^{-1}x+cos^{-1}x=\dfrac{\pi}{2} )\\\\\\=\dfrac{9}{4}sin^{-1}\sqrt{1-(1/2)^2} \\\\\\(cos^{-1}x=sin^{-1}\sqrt{1-x^2} ;0\leq x\leq1)\\ \\\\=\dfrac{9}{4}sin^{-1}(\sqrt{\dfrac{9-1}{9} })\\ \\\\=\dfrac{9}{4}sin^{-1}(\dfrac{2\sqrt{2} }{3} )=R.H.S.$