Question

SecA (1-sinA) (secA+tanA) = 1
Pls solve

Answer 1

To proove : secA ( 1 - sinA ) ( secA + tanA ) = 1



LHS = >


= > secA( 1 - sinA ) ( secA + tanA )


$= > (secA- secAsinA ) \times ( secA + tanA ) \\ \\ \\ = > (secA- ( \frac{1}{cosA} \times sinA )) \times ( secA + tanA ) \\ \\ \\ \\ = > (secA- tan A ) \times ( secA + tanA )$



By formula,
( a + b ) ( a - b ) = a² - b²



= > ( secA )² - ( tanA )²


= > sec²A - tan²A


$= > 1 \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: identity : [ sec^{2}A - tan^{2}A = 1 ] }$




1 = 1
RHS = LHS


Hence, Proved that SecA (1-sinA) (secA+tanA) = 1

Answer 2

Given Equation is secA(1 - sinA)(secA + tanA)

$= > (\frac{1}{cosA})(1 - sinA) * (\frac{1}{cosA} + \frac{sinA}{cosA})$

$= > \frac{1 - sinA}{cosA} * \frac{1 + sinA}{cosA}$

$= > \frac{(1 - sinA)(1 + sinA)}{cos^2A}$

$= > \frac{1 - sin^2A}{cos^2A}$

$= > \frac{cos^2A}{cos^2A}$

= > 1.

Hope it helps!