Question

(1+tan²A) ÷ (1-tan²A) is equal to sec2A
How?

Answer 1

Answer:

=cos A /1+sin A +1+sin A /cos A

=cos ^2A+(1+sinA)^2 /cos A (1+sinA)

= cos^2A+1+sinA+sin ^2A/cosA(1+sin A)

= sin^2A+cos^A+1+2sinA/cosA(1+sinA)

=1+1+2sinA/cosA(1+sinA)

=2+2sinA /cosA(1+sinA)

=2(1+sinA)/cosA(1+sinA)

=2/cosA

=2×1/cosA

=2 sec A

Hence the result

2 sec A=2 sec A

LHS=RHS