Use Cauchy’s mean value theorem to prove that:
{Cos(alpha)- cos(beta) }/{sin(alpha) -sin(beta) }=tan(theta)
Use Cauchy's Mean Value Theorem to prove the given result.
Let "f(x)=sinx,g(x)=cosx" for all "x\\in \\left [ \\alpha ,\\beta \\right ]\\subset \\left [ 0,\\frac{\\pi }{2} \\right ]"
We know that "f(x)" and "f(x)" are derivable on "\\left [ \\alpha ,\\beta \\right ]\\subset R"
And "f(x)=sinx,g(x)=cosx\\Rightarrow f'(x)=cosx,g'(x)=-sinx\\neq 0\\forall x\\in \\left [ \\alpha ,\\beta \\right ]"
By Cauchy's Mean Value theorem there exists "\\theta \\in \\left ( \\alpha ,\\beta \\right )" so that
"\\frac{f(\\beta )-f(\\alpha )}{g(\\beta )-g(\\alpha )}=\\frac{f'(\\theta )}{g'(\\theta )}"
"\\Rightarrow \\frac{sin\\beta -sin\\alpha }{cos\\beta -cos\\alpha }=\\frac{cos\\theta }{-sin\\theta }"
"\\Rightarrow \\frac{cos\\beta -cos\\alpha }{sin\\beta -sin\\alpha }=-\\frac{sin\\theta }{cos\\theta }"
"\\Rightarrow \\frac{cos\\beta -cos\\alpha }{sin\\beta -sin\\alpha }=-tan\\theta"
"\\Rightarrow \\frac{cos\\alpha -cos\\beta }{sin\\alpha -sin\\beta }=-tan\\theta"
Therefore
"\\Rightarrow \\frac{cos\\alpha -cos\\beta }{sin\\alpha -sin\\beta }=-tan\\theta"
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