Question

1+tan2A =sec2A prove that
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Answer 1

Answer:

Here is your solution. Hope it helps

Answer 2

Step-by-step explanation:

Prove that :1+tan²A=sec²A

LHS:1+tan²A

$= 1 + \frac{ {sin}^{2} a}{ { \cos}^{2} \alpha }$

$= \frac{ { \cos }^{2} \alpha + { \sin }^{2} \alpha }{ { \cos}^{2} \alpha }$

$= \frac{1}{ { \cos }^{2} \alpha } ( { \cos}^{2} \alpha + { \sin }^{2} \alpha = 1)$

=sec²A=RHS