Question

sec(x)-cosec(x)=4/3
find x

Answer 1

Convert sec and cosec in the form of sin and cos.

Answer 2

Value of x is $-\frac {1}{2}$

Step-by-step explanation:

Given,  

$sec(x) + cosec(x) = \frac {4}{3}$

We know that $sec(x) = \frac {1}{cos(x)}\ and\ cosec(x) = \frac {1}{sin(x)}$

So,

$\frac {1}{cos(x)} - \frac {1}{sin(x)} = \frac {4}{3}$

Taking LCM and solving we get -

$\frac {sin(x) - cos(x)}{sin(x) \times cos(s)} = \frac {4}{3}$

Dividing both sides by 2 we get-

$\frac {sin(x) - cos(x)}{-2 sin(x) \times cos(s)} = \frac {4}{3(-2)}$

Subtracting with adding, 1 on left side denominator we get-

$\frac {sin(x) - cos(x)}{1-2 sin(x) \times cos(s)-1} = \frac {4}{-6}$

= $\frac {sin(x) - cos(x)}{sin^2(x) + cos^2(x)-2 sin(x)cos(s)-1} = \frac {4}{-6}$

=  $\frac {sin(x) - cos(x)}{(sin(x) + cos(x))^2} = \frac {2}{-3}$

Let us take (sin(x) + cos(x)) = a

So,

$\frac {p}{p^2-1} = \frac {2}{-3}$

or $-3p = 2p^2 -2$

$2p^2 + 3p -2 = 0\ or\ 2p^2 + 4p -p -2 = 0$ (Mid Term Splitting)

(2p-1)(p+2) = 0

Hence, $p = -\frac {1}{2} , -2$

sin(x) - cos(x) can never be -2 (if sin(x) = -1, cos(x) not equals 1)

Hence, $sin(x) - cos(x) = -\frac {1}{2}$