1\int \left(1 + tan x . tan ( x + A ) dx \right)
question is as mentioned by SR (SEE DOWN)
ANS ----> cot A { log sec(x+A) - log sec x}
-
UP 0 DOWN 0 1 5
5 Answers
1there is no x besides tanx ...
since ans is same as given by manmay
1\int \left( 1 + \frac{sin x . sin ( x + A )}{cosx . cos ( x + A) } \right)
\int \left( \frac{cosx . cos ( x + A ) + sin x .sin ( x + A) }{cosx . cos ( x + A ) } \right)dx
\int \left( \frac{cos ( x + A- x ) }{cosx . cos ( x + A ) } \right)dx
cos A \int \left( \frac{dx }{cosx . cos ( x + A ) } \right)
further simplify
1cot A \int \frac{sin A . dx}{cosx . cos ( x + A ) }
cot A \int \frac{sin ( x + A - x) . dx}{cosx . cos ( x + A ) }
cot A \int \left( \frac{sin ( x + A ) . cos x}{cosx . cos ( x + A ) }- \frac{cos ( x + A ) .sin x}{cosx . cos ( x + A ) }\right )dx
