Question

how i can solve dx/(asinx+bcosx)hall square

Answer 1

Hi dear friend,
+++++++++++++++++++++++++++++++
We know the trigonometric identity
Sin²A + cos² A = 1 
-----------------------------------------
acos X + bsinX = m -----( 1 )
asinX - bcosX = n -------( 2 )
Do the square of equations ( 1 ) 
and ( 2) ,
And add resulting equations
a² cos² x+b² sin²x+2abcosxsinx + 
a²sin² x + b² cos²x -2abcosxsinx 
= m² + n²
a² (cos² x + sin² x ) + b² (sin² x + cos² x ) = m² + n²
( From above Trigonometric identity )
a² + b² = m² + n²
Hence proved .
I hope this helps you :)