Question

prove that( 1+tanA-secA) ×(1+tanA+secA)=2tanA
​

Answer 1

Answer:

Step-by-step explanation:

→lets multiply

1+tanA+secA+tanA+tan^2+tanAsecA-secA-secAtanA-secA^2

→let cancel same terms

1+2tanA+tan^2A-sec^2A

(∵we know that sec^2A-tan^2A=1 then multiply with -1 →tan^2A-sec^2A=-1)

1+2tanA-1=2tanA

(∴hence proved)

Answer 2

SOLUTION

TO PROVE

$\sf (1 + \tan A - \sec A)(1 + \tan A + \sec A) = 2\tan A$

FORMULA TO BE IMPLEMENTED

$\sf { \sec }^{2} A - {\tan}^{2} A = 1$

EVALUATION

LHS

$\sf = (1 + \tan A - \sec A)(1 + \tan A + \sec A)$

$\sf = {(1 + \tan A)}^{2} - {( \sec A) }^{2}$

$\sf =1 + 2\tan A +{\tan}^{2} A - { \sec }^{2} A$

$\sf =1 + 2\tan A - { \sec }^{2} A+{\tan}^{2} A$

$\sf =1 + 2\tan A - ({ \sec }^{2} A - {\tan}^{2} A )$

$\sf =1 + 2\tan A - 1$

$\sf = 2 {\tan}^{} A$

= RHS

Hence proved

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