Question

prove that (tanA+secA-1) / (tanA-secA+1) = secA+tanA

Answer 1

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Answer 2

Answer:

The proof is given below:

Step-by-step explanation:

Given,

$\frac{tanA+secA-1}{tanA-secA+1}$

Now we know,

sec²A - tan²A = 1

Substituting this identity,

= $\frac{tanA+secA-(sec^{2}A-tan^{2}A)}{tanA+1-secA}$

= $\frac{tanA+secA-((secA+tanA)(secA-tanA))}{tanA+1-secA}$

= $\frac{(tanA+secA)(1-(secA-tanA)}{tanA+1-secA}$

= $\frac{(tanA+secA)(1-secA+tanA)}{tanA+1-secA}$

= secA+tanA

Hence proved