Question

prove that secA+tanA-1 × secA-tanA+1=2tanA

Answer 1

We can write the above equation as{(secA)+(tanA-1)}*{(secA) - (tanA-1)}It is in the formula (a+b)*(a-b) = a²-b²So,=> (secA)² - (tanA-1)²=> sec²A - [tan²A + 1 - 2tanA]=> sec²A - tan²A - 1 +2tanA=> (sec²A - tan²A) - 1 +2tanA ......As sec²A - tan²A = 1So,=> 1 - 1 +2tanA= 2tanA
There is an alternate method too:

[secA+(tanA-1)(secA-(tanA-1)]
=sec²A-(tanA-1)²
=1/cos²A-(tan²A-2tanA+1)
=1+tan²A-tan²A+2tanA-1
=(1-1)tan²A-tan²A + 2tanA
=0+0+2tanA
=2tanA

Hope This Helps :)

Answer 2

Ans.

[secA +(tan A-1)] × [secA -(tan A- 1)]

° we know that
(a² - b²) = (a+b) (a-b)

so

sec²A - (tanA-1)²

1 + tan²A - (tan²A + 1 - 2tanA)

1 + tan²A - tan²A - 1 + 2tanA

= 2tanA

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