Question

cos 9A=sinA then ,tan 5 A= ?​

Answer 1

Here tanA=cos9A/sin9A=cot9A
Now tan5A=tan(4A+A)=(tan4A+tanA)/1-(tan4AtanA)

Now tan4A=tan(2A+2A)=2tan2A/1-tan^2(2A)

Now tan2A=2tanA/1-tan^2(A)
So putting values of tanA in reverse direction you get your answer