Question

prove that sec A - tan A = 1/sec A+tan A​

Answer 1

To prove:-

$</p><p>   </p><p>\tt {\small {\red {sec^A - tan^A = \frac{1}{sec A + tan A}}}$ \\

Proof:-

$</p><p></p><p>\tt sec A - tan A \: = \: LHS \\ \\</p><p></p><p> \tt \frac{sec A - tan A}{1} \times (\frac{sec A + tan A}{sec A + tan A}) \\</p><p></p><p>\tt ........(rationalizing \:  the \: denominator) \\ \\</p><p></p><p>\tt \therefore \frac{sec^2  - tan^2 A}{sec A + tan A} \\ \\</p><p></p><p>\tt \therefore \frac{1}{sec A + sec A } \\ </p><p></p><p>\tt .....( since \: 1 + tan^2 A = sec^2 A ) \\ \\</p><p></p><p>\tt \therefore \frac{1}{sec A + tanA } = RHS \\ \\ </p><p></p><p>\tt \therefore \: LHS \: + \: RHS \\ \\</p><p></p><p>\tt \orange {Hence \: proved \:  that, } \\</p><p></p><p>\tt sec A - tan A = \frac{1}{sec A + tan A} \\ \\ \\</p><p></p><p>\sf {More \: Information:-} \\ \\</p><p></p><p>\tt sin^2A = cos^2A = 1 \\ \\</p><p></p><p>\tt 1 + tan^2 A = sec^2 A \\ \\</p><p></p><p>\tt 1 + cot^2 A = cosec^2A \\ \\ </p><p></p><p>\tt These \: are \: fundamental\:  trigonometric \: Identities \\</p><p></p><p>\tt helpful \: in \: solving \: trigonometric \:problems \\</p><p>$

Answer 2

Step-by-step explanation:

we know that,

=>sec^2A-tan^2A=1

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

{ since ,(a+b)(a-b)=a^2-b^2 }

=>secA - tanA =1/ sec A + tan A

hence ,proved.