If asin³X + bcos³X = sinXcosX and asinX = bcosX, then find the value of a² + b², provided that X is neither 0° nor 90°.

Correct option is B

​Given:

​$$a \sin^3 X + b \cos^3 X = \sin X \cos X$$​​

​$$a \sin X = b \cos X$$​​

X is neither 0° nor 90°

Formula Used:

​$$\sin^2 X + \cos^2 X = 1$$ 

​$$​​a \sin X = b \cos X \implies a = b \frac{\cos X}{\sin X} ​$$​​

Solution:

Substituting $$a = b \frac{\cos X}{\sin X}$$​ into the first equation

​$$a \sin^3 X + b \cos^3 X = \sin X \cos X:$$​

= $$b \frac{\cos X}{\sin X} \sin^3 X + b \cos^3 X = \sin X \cos X$$​

= $$b \cos X \sin^2 X + b \cos^3 X = \sin X \cos X$$​​

​$$\implies b \cos X (\sin^2 X + \cos^2 X) = \sin X \cos X$$​​

Since $$\sin^2 X + \cos^2 X = 1$$​, So:

​$$b \cos X = \sin X \cos X$$​​​

​$$b = \sin X$$​​

Substituting $$b = \sin X$$​ into $$a \sin X = b \cos X$$​:

​$$a \sin X = \sin X \cos X$$​​

So, $$a = \cos X.$$​​

Now:

​$$a^2 + b^2 = \cos^2 X + \sin^2 X = 1$$​​​​