Science:Math Exam Resources/Courses/MATH102/December 2014/Question B 03/Solution 1

In this problem we know the point is at $a={\frac {\pi }{4}}$ and the function value $f(a)=f({\frac {\pi }{4}})=\tan({\frac {\pi }{4}})=1$. To calculate $f'({\frac {\pi }{4}})$ we need the derivative of $f(x)$, which is $f'(x)=(\tan(x))'=\sec ^{2}(x),$ and it follows $f'({\frac {\pi }{4}})=\sec ^{2}({\frac {\pi }{4}})=2$. So the equation of the tangent line is $L(x)=1+2(x-{\frac {\pi }{4}}).$