Question

Find the value sin(cos^-1 4/5 + tan^-1 2/3)​

Answer 1

Answer:

$sin(cos^{-1}\frac{4}{5}+ tan^{-1}\frac{2}{3})=\frac{17}{5\sqrt{13}}$

Step-by-step explanation:

Formula used:

sin(A+B) = sinA cosB + cosA sinB

$Take\\\\cos^{-1}\frac{4}{5}=A\\\\cosA=\frac{4}{5}\\\\sin^2A=1-cos^2A\\\\sin^2A=1-\frac{16}{25}\\\\sin^2A=\frac{9}{25}\\\\sinA=\frac{3}{5}$

$Take\\\\\\tan^{-1}\frac{2}{3}=B\\\\tanB=\frac{2}{3}\\\\sec^2B=1+tan^2B\\\\sec^2B=1+\frac{4}{9}\\\\sec^2B=\frac{13}{9}\\\\cos^2B=\frac{9}{13}\\\\cosB=\frac{3}{\sqrt{13}}$

$sin^2B=1-cos^2B\\\\sin^2B=1-\frac{9}{13}\\\\sin^2B=\frac{4}{13}\\\\sinB=\frac{2}{\sqrt{13}}$

$we\:have \\\\sinA=\frac{3}{5}\\\\sinB=\frac{2}{\sqrt{13}}\\\\cosA=\frac{4}{5}\\\\cosB=\frac{3}{\sqrt{13}}$

$Now\\\\sin(cos^{-1}\frac{4}{5}+ tan^{-1}\frac{2}{3})\\\\=sin(A+B)\\\\=sinA\:cosB+cosA\:sinB\\\\=\frac{3}{5}.\frac{3}{\sqrt{13}}+\frac{4}{5}.\frac{2}{\sqrt{13}}\\\\=\frac{9}{5\sqrt{13}}+\frac{8}{5\sqrt{13}}\\\\=\frac{17}{5\sqrt{13}}$

Answer 2

Answer:

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