Question

sec A - tanA / sec A + tan A = 1-2sec A tan A + 2 tan^2 A
​

Answer 1

To Prove:

$\frac{secA-tanA}{secA+tanA} =1-2secAtanA+2tan^{2} A$

Solution:

We know that,

• (a-b)(a-b)= (a-b)²
• (a+b)(a-b)= a²-b²
• sec²x - tan²x= 1 ⇒ 1+tan²x=sec²x
• (a-b)²= a²+b²-2ab

$\sf{L.H.S.=\dfrac{secA-tanA}{secA+tanA}}$

[Rationalising]

       $\implies\sf\dfrac{secA-tanA}{secA+tanA}\times$  $\sf\dfrac{secA-tanA}{secA-tanA}$

        $\implies\sf\dfrac{(secA -tanA)^{2} }{sec^{2}A -tan^{2} A}$

        $\implies\sf\dfrac{sec^{2} A +tan^{2}A-2secAtanA }{1}$

        $\implies\sf{1+tan^{2} A +tan^{2}A-2secAtanA}$

        $\implies\sf{1+2tan^{2} A -2secAtanA}$

        $\implies\sf{1 -2secAtanA+2tan^{2}A}$

         $\implies\sf{R.H.S.}$

Hence Proved