Question

sin-1[sinx+cosx]/root2

Answer 1

Answer:

$\red {sin^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)}\green {= x + \frac{\pi}{4}}$

Step-by-step explanation:

$sin^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)$

$= sin^{-1}\left( \frac{1}{\sqrt{2}}\times sinx + \frac{1}{\sqrt{2}}\times cosx\right)$

$= sin^{-1}\left(cos \frac{\pi}{4}\times sinx + sin \frac{\pi}{4}\times cosx\right)$

$= sin^{-1}[sin(x + \frac{\pi}{4} )]$

$\boxed {\pink { SinAcosB + sinBcosA = sin(A+B)}}$

$= x + \frac{\pi}{4}$

$\boxed { \pink {Since, \:sin^{-1}(sinx) = x }}$

Therefore.,

$\red {sin^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)}\green {= x + \frac{\pi}{4}}$

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Answer 2

Answer:

1 it is correct answer please mark my answer as Brainlist