Question

prove that secA+tanA/secA-tanA= 1+2sec A×tanA+2tan²A please be fast​

Answer 1

$\sf{Multiply \: and \: divide \: by \: secA + tanA}$

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

$\implies \sf{ \frac{sec {}^{2} A + tan {}^{2} A + 2secAtanA}{1}} \\ \\ \implies \sf{(1 + tan {}^{2} A) + tan {}^{2} A + 2secAtanA } \\ \\ \implies\sf{1 + 2tan {}^{2} A + 2secAtanA}$

$\sf{Used \: trigonometric \: properties: }$

• sec²A - tan²A = 1

• sec²A = 1 + tan²A