`(sin A+cos A)/(sin A-cos A)` . Solve, if tan A = 6

(cos A + sin A) / (cos A-sin A) - (cos A-sin A) / (cos A + sin A) = 2tan2A

(1 + cot A + tan A) (sin A-cos A) = sin A tan A-cot A cos A

If A lies in the first quadrant and 6 tan A=5, then the value of ( 8 sin A - 4 cos A )/( cos A + 2 sin A) is :

(sin A + sin2A) / (1 + cos A + cos2A) = tan A

If (sin x + cos x)/(sin x - cos x) = (6)/(5) , then the value of (tan^(2) x + 1)/(tan^(2) x-1) is :

(sin ^ (2) A-sin ^ (2) B) / (sin A cos A-sin B cos B) = tan (A + B)

Prove that (i) (sec 8A -1)/(sec 4A -1) =(tan 8A)/(tan 2A) (ii) (cos A+ sin A)/(cos A - sin A)-(cos A-sin A)/(cos A+sin A)=2 tan 2A

(sin A+sin3A)/(cos A+cos3A)=tan2A

(sin A sin2A + sin3A sin6A) / (sin A cos2A + sin3A cos6A) = tan5A

(sin A sin2A + sin3A sin6A + sin4A sin13A) / (sin A cos2A + sin3A cos6A + sin4A cos13A) = tan9A