The correct option is A tan A√1+tan2A,1√1+tan2A
We know that
tan2A+1=sec2A⟹secA=√tan2A+1.
As secA and cosA are reciprocals of each other,
we can directly write the value of cosA as:
cos A=1√1+tan2A and from the identity sin2A+cos2A=1
we can get the value of sinA.
sinA=√1−cos2A.
Substitute value of cosA=1√1+tan2A in the above equation.
sinA=√1−(1√1+tan2A)2
=√1−11+tan2A
=√1+tan2A−11+tan2A
=√tan2A1+tan2A
⟹sinA=tanA√1+tan2A and cosA=1√1+tan2A