Question

if secA-tanA=√2tanA then show that secA+tanA=√2secA

Answer 1

here is the answer........

Answer 2

$Given \:Sec A - tan A = \sqrt{2} tan A \: ---(1)$

/* On squaring both sides ,we get */

$\implies (Sec A - tan A)^{2} = (\sqrt{2} tan A)^{2}$

$\implies Sec^{2} A + tan^{2} A - 2sec A tan A = 2 tan^{2} A$

$\implies Sec^{2} A - 2sec A tan A = 2 tan^{2} A-tan^{2} A$

$\implies Sec^{2} A - 2sec A tan A = tan^{2} A$

$\implies Sec^{2} A = tan^{2} A + 2SecA tan A$

$\implies Sec^{2} A + \pink {Sec^{2} A} = tan^{2} A + 2SecA tan A + \pink {Sec^{2} A}$

$\implies 2Sec^{2} A = (tan A + SecA )^{2}$

$\implies \sqrt{2Sec^{2} A }= \sqrt{(tan A + SecA )^{2} }$

$\implies \blue {\sqrt{2}Sec A = tan A + SecA }$

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