Question

If secA+tanA=x,prove that sinA=x^2-1/x^2+1.

Answer 1

$Given \:secA + tan A = x \: --(1)$

/* Multiplying both sides of the equation by

(sec A - tan A ) , we get */

$\implies (secA+tanA)(secA-tanA) = x(secA-tanA)$

$\implies sec^{2} A - tan^{2} A = x(secA-tanA)$

$\implies 1 = x(secA-tanA)$

$\boxed {\pink {Since ,sec^{2}A - tan^{2} A = 1 }}$

$\implies \frac{1}{x} = secA - tan A \: --(2)$

$Subtract equation (2) from equation (1) , we get */</p><p>[tex] \implies 2tan A = x - \frac{1}{x}$

$\implies 2tan A = \frac{(x^{2} - 1)}{x}\: --(3)$

/* Adding equation (1) and equation (2) , we get */

$\implies 2secA = x + \frac{1}{x}$

$\implies 2sec A = \frac{(x^{2} + 1)}{x}\: --(4)$

$Now, Do \: Equation\: (3) \div Equation\: (4) ,\\ we \: get$

$\implies \frac{2tanA}{2secA } = \frac{\frac{x^{2}-1}{x}}{\frac{x^{2}+1}{x}}$

$\implies \frac{\frac{sin A}{cosA}}{\frac{1}{cos A }} = \frac{x^{2}-1}{x^{2}+1}$

$\implies sinA = \frac{x^{2}-1}{x^{2}+1}$

$Hence , proved .$

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