Question

(secA+tanA_ 1)( secA_ tanA+ 1)= 2tanA

Answer 1

Hello friend

------------------------------------------------------------------------------------------------------------
Given
(sec A+tanA-1)( secA-tanA+ 1)= 2tanA

LHS  (sec A+tanA-1)( secA-tanA+ 1)
=> sec²A-secAtanA+secA+tanAsecA-tan²A+tanA-secA+tanA
=> sec²A-tan²A+2tanA-1
[ ∴ sec²A-tan²A=1]
=> 1+2tanA-1
=2tanA
RHS = 2tanA

Hence proved ∴ LHS=RHS

-----------------------------------------------------------------------------------------------------------------------

hope it's help u :))

Answer 2

$(secA + tanA - 1)(secA - tanA + 1) \\ \\ = ( \frac{1 + sinA - cosA}{cosA} )( \frac{1 - sinA + cosA}{cosA} ) \\ \\ = \frac{ {1}^{2} - ( {sinA - cosA)}^{2} }{ {cos}^{2}A } \\ \\ = \frac{2sinacosA}{ {cos}^{2}A } \\ \\ = 2tanA$