Given $$a\space sin\space q=b\space cos\space q$$​ and $$a\space sin^3\space q+b\space cos^3\space q=sin\space q cos\space q$$​, for $$0<q<90°$$​, find the value of $$a^2+b^2$$​.

Correct option is B

Given:

​$$a\space sin\space q=b\space cos\space q$$​​

​$$a\space sin^3\space q+b\space cos^3\space q=sin\space q cos\space q$$​​

Formula Used:

​$$\sin^2 \theta + \cos^2 \theta =1$$​​

Solution:

​$$a\space sin\space q=b\space cos\space q$$.....(1) 

​$$a\space sin^3\space q+b\space cos^3\space q=sin\space q cos\space q$$...(2)

​$$(a\space sin\space q)(\space sin^2\space q)+b\space cos^3\space q=sin\space q cos\space q$$ 

From eq.(1)

​$$(b\space cos\space q)(\space sin^2\space q)+b\space cos^3\space q=sin\space q cos\space q$$​​

​$$(b\space cos\space q)(\space sin^2\space q+\space cos^2\space q)=sin\space q cos\space q$$​​

​$$(b\space cos\space q)(1)=sin\space q cos\space q$$​​

​$$b=sin\space q$$​​

Then $$a =\frac{b\space \cos \space q}{\sin q} = \cos q$$​

​$$Then \ (a^2 + b^2 ) = \sin^2 q + \cos^2 q =1$$​​