Question

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Answer 1

Answer:

$\sec \: a = \frac{ {x}^{2} + 1}{2x}$

Explanation:

Given:

$sec \: a + \: tan \: a = x \: .............(i)$

Steps:

Method 1:

We can write above equation as below:

$\sec \: a = x - \tan \: a$

Squaring both sides, we get:

We know that,

$(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab$

${ \sec}^{2} \: a = { x}^{2} + { \tan}^{2} \: a - 2 \: x \tan \: a$

${ \sec }^{2} \: a - { \tan }^{2} \: a = {x}^{2} - 2 \: x \tan \: a$

We know that,

${ \sec}^{2} \: a - { \tan }^{2} \: a = 1$

Substituting the value in above equation, we get:

$1 = {x}^{2} - 2 \: x \tan \: a$

We can write the above equation as below:

$1 - {x}^{2} = - 2 \: x\tan \: a$

${x}^{2} - 1 = 2 \: x \tan \: a$

$\frac{ {x}^{2} - 1 }{2x} = \tan \: a$

Substituting the value of tan a in equation (i), we get:

$\sec \: a + \frac{ {x}^{2} - 1 }{2x} = x$

$\sec \: a = x - \frac{ {x}^{2} - 1}{2x}$

$\sec \: a = \frac{2 {x}^{2} - {x}^{2} - 1 }{2x}$

$\sec \: a = \frac{ {x}^{2} - 1}{2x}$

Method 2:

We know that,

${ \sec}^{2} \: a - { \tan }^{2} \: a = 1$

We know that,

${a}^{2} - {b}^{2} = (a + b)(a - b)$

We can write,

${ \sec}^{2} \: a - { \tan}^{2} \: a = ( \sec \: a + \tan \: )( \sec \: a - \tan \: a)$

$( \sec \: a + \tan \: a)( \sec \: a - \tan \: a) = 1$

It is given in the question,

$(\sec \: a + \tan \: a) = x$

Substituting this value in above equation, we get:

$x( \sec \: a - \tan \: a) = 1$

We can write above equation as below:

$\sec \: a- \tan \: a = \frac{1}{x}$

Adding the above equation to (i), we get:

$2 \sec \: a = x + \frac{1}{x}$

Taking LCM and adding, we get:

$2 \sec \: a = \frac{ {x}^{2} + 1 }{x}$

We can write the above equation as below:

$\sec \: a = \frac{ {x}^{2} + 1 }{2x}$

Therefore, the answer is:

$\frac{ {x}^{2} + 1 }{2x}$