#### Question

Prove that the function f given by f(x) = log cos x is strictly increasing on (−π/2, 0) and strictly decreasing on (0, π/2) ?

#### Solution

\[f\left( x \right) = \log \cos x\]

\[f'\left( x \right) = \frac{1}{\cos x}\left( - \sin x \right)\]

\[ = - \tan x\]

\[\text { Now,} \]

\[x \in \left( - \frac{\pi}{2}, 0 \right)\]

\[ \Rightarrow \tan x < 0\]

\[ \Rightarrow - \tan x > 0 \]

\[ \Rightarrow f'(x) > 0\]

\[\text { So,f(x)is strictly increasing on } \left( - \frac{\pi}{2}, 0 \right) . \]

\[\text { Now,} \]

\[x \in \left( 0, \frac{\pi}{2} \right)\]

\[ \Rightarrow \tan x > 0\]

\[ \Rightarrow - \tan x < 0 \]

\[ \Rightarrow f'(x) < 0\]

\[\text { So,f(x)isstrictly decreasing on }\left( 0, \frac{\pi}{2} \right).\]

Is there an error in this question or solution?

Solution Prove that the Function F Given by F(X) = Log Cos X is Strictly Increasing on (−π/2, 0) and Strictly Decreasing on (0, π/2) ? Concept: Increasing and Decreasing Functions.