Question

plz solve this
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Answer 1

(secA -tanA)²

=sec²A + tan²A -2secAtanA

=(1+tan²A) + tan²A -2secAtanA

=1+2tan²A -2sinA/cos²A

=1 + 2sin²A/cos²A  -2sinA/cos²A

=(cos²A + 2sin²A -2sinA)/cos²A

=(1+sin²A -2sinA)/(1-sin²A)

=(1-sinA)²/(1+sinA)(1-sinA)

=(1-sinA)/1+sinA)

Answer 2

Given:

$(secA-tanA)^2$

$\implies sec^2A+tan^2A-2secAtanA$

$\implies sec^2A-tan^2A+2tan^2A-2secAtanA$

We know that $sec^2A-tan^2A=1$

$\implies 1+2tan^2A-2secAtanA$

$\implies 1+\frac{2sin^2A}{cos^2A}-\frac{2sinA}{cos^2A}$

$\implies \frac{cos^2A+2sin^2A-2sinA}{cos^2A}$

$\implies \frac{cos^2A+sin^2A+sin^2A-2sinA}{cos^2A}$

We know that $cos^2A+sin^2A=1$

$\implies \frac{1+sin^2A-2sinA}{cos^2A}$

$\implies \frac{(1-sinA)^2}{1-sin^2A}$

$\implies \frac {(1-sinA)(1-sinA)}{(1+sinA)(1-sinA)}$

$\implies \frac{1-sinA}{1+sinA}$

Hence proved.