Question

Prove: $\frac{secA- tanA}{secA+ tanA} = 1 - 2 sec A. tan A + 2 tan^2A$

Answer 1

check the attachment

Answer 2

$\underline{\huge{\textsf{To Prove:}}}$

$\frac{sec\: A- tan\: A}{sec\: A+ tan\: A} = 1 - 2 sec\: A. tan\: A + 2tan^{2}\: A$

$\underline{\underline{\huge{\textbf{Solution:}}}}$

$\underline{\large{\textsf{Step-1:}}}$
Consider LHS

$\underline{\textbf{LHS=}}$

$\implies \frac{sec\: A- tan\: A}{sec\: A+ tan\: A}$

$\underline{\large{\textsf{Step-2:}}}$
Rationalize the Numerator

i.e., Multiply and Divide with $(sec\: A- tan\: A)$ on both sides

$\implies \frac{sec\: A- tan\: A}{sec\: A+ tan\: A} \times \frac{sec\: A- tan\: A}{sec\: A- tan\: A}$

$\implies \frac{(sec\: A- tan\: A)^{2}}{(sec\: A+ tan\: A)(sec\: A- tan\: A)}$ ————[1]

$\underline{\large{\textsf{Step-3:}}}$
Using the identity $\sf \mathbf{(a-b)^{2} = a^{2} +b^{2} - 2ab}$

$(sec\: A- tan\: A)^{2} =sec^{2}\: A+ tan^{2}\: A- 2sec\: A\, tan\: A$

Substituting in [1]

$\implies \frac{sec^{2}\: A+ tan^{2}\: A- 2sec\: A\, tan\: A}{(sec\: A+ tan\: A)(sec\: A- tan\: A)}$ ————[2]

$\underline{\large{\textsf{Step-4:}}}$
Using the identity $\sf \mathbf{(a+b)(a-b) = a^{2} -b^{2}}$

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

Substituting in [2]

$\implies \frac{sec^{2}\: A+ tan^{2}\: A- 2sec\: A\, tan\: A}{sec^{2}\: A- tan^{2}\: A}$ ————[3]

$\underline{\large{\textsf{Step-5:}}}$
Using the identity $\sf \mathbf{sec^{2}\: \theta = 1 + tan^{2}\: \theta}$

$sec^{2}\: A = 1+tan^{2}\: A$
Substituting in [3]

$\implies \frac{(1+tan^{2}\: A)+ tan^{2}\: A- 2sec\: A\, tan\: A}{(1+tan^{2}\: A)- tan^{2}\: A}$

$\implies \frac{1+2 tan^{2}\: A- 2sec\: A\, tan\: A}{1}$

$\implies 1+2 tan^{2}\: A- 2sec\: A\, tan\: A$

$\implies 1- 2sec\: A\, tan\: A +2 tan^{2}\: A$

$\underline{\textbf{=RHS}}$

$\textsf{Hence Proved}$