Question

if tanA+secA=m then prove that secA=m^2+1/2 ​

Answer 1

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TO ProvE

$\implies \: \boxed{ tanA + secA = m}...........(1) \\ now \: we \: know \: that \\ \implies \boxed{ \red{sec {}^{2} A - tan {}^{2} A = 1}} \\ \implies \: (secA + tanA)(secA - tanA) = 1 \\ \implies \: (m) (secA - tanA) = 1 \\ \implies \: \boxed{(secA - tanA) = \frac{1}{m}}..........(2) \\ now \: doing \:( equ {}^{n} (1) + equ {}^{n} (2))...we \: get \\ \implies \:(tanA + secA) + (secA - tanA) = 1 + \frac{1}{m} \\ \implies \: tanA + secA + secA - tanA = \frac{m + 1}{m} \\ \implies2secA = \frac{m + 1}{m} \\ \implies \: secA = \frac{1 + m}{2m}$

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