Question

sec(x) - cosec(x) = 4/3 then what is sin(x) -cos(x)

Answer 1

As we know,
Sec(x) = $\frac{1}{cos(x)}$
Cosec(x) = $\frac{1}{Sin(x)}$
Hence, we will convert given equation in form of cos(x) and sin(x)
sec(x) - cosec(x) = 4/3
=>$\frac{1}{cos(x)}$ - $\frac{1}{sin(x)}$ = 4/3
=>$\frac{sin(x) - cos(x)}{sin(x)*cos(x)} = \frac{4}{3}$
Multiplying denominator both sides by -2,
=>[tex] \frac{sin(x) - cos(x)}{-2cos(x).sin(x)} = \frac{4}{3(-2)} [/tex]
Adding and subtracting 1 from left side denominator,
$\frac{sin(x)-cos(x)}{1-2sin(x)cos(x)-1} = \frac{4}{-6}$
=>$\frac{sin(x)-cos(x)}{ sin^{2}(x) + cos^{2}(x) -2sin(x)cos(x)-1} = \frac{4}{-6}$
=> $\frac{sin(x)-cos(x)}{ (sin(x)-cos(x))^{2} -1} = \frac{2}{-3}$
Let us assume (sin(x)-cos(x) = p
=>$\frac{p}{ (p)^{2} -1} = \frac{2}{-3}$
=>$-3p =2p^{2} -2$
=>$2p^{2} +3p -2= 0 =\ \textgreater \ 2p^{2} +4p -p -2 = 0 =\ \textgreater \ 2p(p+2) -1(p+2)=0$
=>(2p+1)(p+2) = 0
=> p = -$\frac{1}{2}$, -2
sin(x) - cos(x) can never be -2 (if sin(x) = -1, cos(x) $\neq$ 1)
Hence, sin(x)-cos(x) = -$\frac{1}{2}$