Question

y=sin^-1[(asinx + bcosx)/✓a^2+b^2] then dy/dx​

Answer 1

Answer:

This is your answer

Step-by-step explanation:

Step-by-step explanation:

y = Sin⁻¹ (a + bCosx)/(b + a Cosx)

=> Siny = (a + bCosx)/(b + a Cosx)

=> Cosy * dy/dx =   (a + bCosx) (aSinx)/(b + a Cosx)²   + (-bSinx)/(b + a Cosx)

=> Cosy * dy/dx =  ( (a + bCosx) (aSinx) + (b + a Cosx)(-bSinx))/(b + a Cosx)²

=>  Cosy * dy/dx =  ( a²Sinx + abCosxSinx -abCosxSinx - b²Sinx))/(b + a Cosx)²

=>  Cosy * dy/dx =  ( a²Sinx  - b²Sinx))/(b + a Cosx)²

=>  Cosy * dy/dx =  -Sinx( b² - a²))/(b + a Cosx)²

Siny = (a + bCosx)/(b + a Cosx)

=> Cos²y = 1  - ((a + bCosx)/(b + a Cosx))²

=> Cos²y = (b² + a²Cos²x + 2abCosx  - (a² + b²Cos²x + 2abCosx) )/(b + a Cosx)²

=> Cos²y = (b²sin²x - a²Sin²x )/(b + a Cosx)²

=> Cosy = Sinx √(b² - a²)/(b + a Cosx)

(Sinx √(b² - a²)/(b + a Cosx)) * dy/dx =   -Sinx( b² - a²))/(b + a Cosx)²

=> √(b² - a²) * dy/dx = - ( b² - a²))/(b + a Cosx)

=> dy/dx = - ( √(b² - a²))/(b + a Cosx)

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