Question

If sec A + tan A = x, then find tan A.​

Answer 1

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Solution:

Given: sec A+ tan A = x (1)

We have

sec² A tan² A = 1

$\sf \: → (sec A+ tan A) (sec A- tan A) = 1 \: \: [a² - b² = (a + b)(a - b)]$

$\sf \: \sec(A) - \tan(A) = \frac{1}{x} →(2)$

$\sf \: Subtracting \: the \: equation \\ \sf \: (2) \: from \: (1) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: ⇒ sec A + tan A- (sec A- tan A) = x \: - \: \frac{1}{x}$

$\sf \: ⇒ sec A+ tan A - sec A+ tan A) = \frac{ {x}^{2} - 1 }{x}$

$\sf \: ⇒ 2 \: tan A = \frac{ {x}^{2} - 1 }{x}$

$\sf \: ⇒tan A = \frac{ {x}^{2} - 1 }{2x}$

hope it was helpful to you

Answer 2

Given:

Sec A + tan A = x

To find:

Value of tan A

Solution:

By trignometric property,

$Sec^{2}A=tan^{2}A+1$

$Sec^{2}A-tan^{2}A=1$

$Sec^{2}A-tan^{2}A$ is of the form $a^{2}-b^{2}=(a+b)(a-b)$

$(Sec A + tan A)(Sec A-tan A) = 1...(1)$

$SecA+tanA=x...(2)$  (given)

Substituting equation (2) in equation (1)

$x(SecA-tanA)=1$

$Sec A-tan A=\frac{1}{x}$

$tanA=SecA+\frac{1}{x}$

Value of tan A is given by, $tanA=SecA+\frac{1}{x}$