trigo !!!!

now log(cosθ) will always be negative in the interval
{infact log(cosθ)is always negative except when θ=0 when it becomes 0}

now for cos(logθ) we notice that cos(log(e^{-Π/2})=0
and all the values after it will be positive till θ<Π/2

so cos(logθ)>log(cosθ)

for cos (log x) to exist. xmust be betn 0 and ∩/2.

taking that condn log x in that region is very close to 0. so cos(logx) is very close to 1.as cosine graph decreases from 1 to 0the interval 0-∩/2.

cos x in the region is betn0-1.so log(cos x) is negative in the region as log of something betn 0 and 1 with base>1 is negative

so cos(log x)>log(cos x)