Question

secA = x + 1/4x
Then prove that secA + tanA = 2x or 1/2x

Please type neatly or provide written answers for clarity

Answer 1

Solution :-

secA = x + 1/4x

Squaring on both sides

⇒ sec²A = (x + 1/4x)²

Since sec²A = tan²A + 1

⇒ tan²A + 1 = (x + 1/4x)²

⇒ tan²A = (x + 1/4x)² - 1

⇒ tan²A = (x + 1/4x)² - 4(x)(1/4x)

⇒ tan²A = (x - 1/4x)²

Since (a + b)² - 4ab = (a - b)²

Taking square root on both sides

⇒ √tan²A = √(x - 1/4x)²

⇒ tanA = ±(x - 1/4x)

⇒ tanA = +(x - 1/4x) or tanA = -(x - 1/4x)

⇒ tanA = x - 1/4x or tanA = - x + 1/4x

For tanA = x - 1/4x

secA + tanA

= x + 1/4x + (x - 1/4x)

= x + 1/4x + x - 1/4x

secA + tanA = 2x

For tanA = - x + 1/4x

secA + tanA

= x + 1/4x + (- x + 1/4x)

= x + 1/4x - x + 1/4x

= 2/4x

secA + tanA = 1/2x

So SecA + tanA = 2x or 1/2x

Hence proved

Answer 2

Solution:

Given:

$\sf{\implies \sec A = x + \dfrac{1}{4x}}$

To prove:

$\sf{\implies \sec A + \tan A = 2x\;or\;\dfrac{1}{2x}}$

So,

$\sf{\implies \sec A = x + \dfrac{1}{4x}}$

Squaring both sides, we get

$\sf{\implies \sec^{2} A = \bigg(x + \dfrac{1}{4x}\bigg)^{2}}$

We know that sec² A = 1 + tan² A.

$\sf{\implies 1+ \tan^{2} A  = \bigg(x+\dfrac{1}{4x}\bigg)^{2}}$

$\sf{\implies \tan^{2} A = \bigg(x + \dfrac{1}{4x}\bigg)^{2} - 1}$

$\sf{\implies \tan^{2}A = x^{2} + \dfrac{1}{16x^{2}} + \dfrac{1}{2}-1}$

$\sf{\implies \tan^{2} A = x^{2}+\dfrac{1}{16x^{2}}-\dfrac{1}{2}}$

$\sf{\implies \tan^{2} A = \bigg(x - \dfrac{1}{4x}\bigg)^{2}}$

$\sf{\implies \tan A = x - \dfrac{1}{4x} \; or \; -\bigg(x-\dfrac{1}{4x}\bigg)}$

When we put tan A = x - 1/4x, we get

$\sf{\sec A + \tan A = x + \dfrac{1}{4x} + x - \dfrac{1}{4x} = 2x}$

When we put tan A = -(x - 1/4x), we get

$\sf{\sec A + \tan A = x + \dfrac{1}{4x}-\bigg(x-\dfrac{1}{4x}\bigg)}$

$\sf{= x + \dfrac{1}{4x} - x + \dfrac{1}{4x}}$

$\sf{\implies \dfrac{2}{4x}= \dfrac{1}{2x}}$

Hence Proved!!