Question

use cauchy's mean value theorem to prove that {cosa-cosB/sina-sinB}=tan0,0<a<0<B<π/2

Answer 1

Let f(x) = sin x and g(x) = cos x

Then, f'(x) = cos x and g'(x) = -sin x

Clearly, f and g are continuous on [a,B]

and derivable on (a,B)

Now using cauchy's mean value theorem,

(f(B)-f(a))/(g(B)-g(a)) = f'( $\theta$ )/g'($\theta$ )

$\forall$ $\theta$$\isin$ (a,B)

(sin B - sin a)/(cos B - cos a) = cos θ /-sin θ

By taking reciprocal on both sides,

(cos a - cos B)/(sin a - sin B) = tan $\theta$