Math, asked by mohdnaveed, 1 year ago

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

Answers

Answered by KhushbooBhaskar
9
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 <br />=\ \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}

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