Question

Find the value of
$\left[\cos 1-\cos ^{-1} 1\right]-\left[\sin 1-\sin ^{-1} 1\right]+\left[\tan 1-\tan ^{-1} 1\right]-\left[\cot 1-\cot ^{-1} 1\right]+\left[\sec 1-\sec ^{-1} 1\right]-\left[\operatorname{cosec} 1-\operatorname{cosec}^{-1} 1\right]$
where $[x]$ denotes greatest integer less than or equal to $x$.

Answer 1

As $ \cos 1-\cos ^{-1} 1=\cos 1-0 \in(0,1) \Rightarrow \left[\cos 1-\cos ^{-1} 1\right]=0$
$\sin 1-\sin ^{-1} 1=\sin 1-\frac{\pi}{2} \in(-1,0) \Rightarrow\left[\sin 1-\sin ^{-1} 1\right]=-1 $
$\tan 1-\tan ^{-1} 1=\tan 1-\frac{\pi}{4} \approx 1.73-0.78>0 \text { and } \in(0,1) \Rightarrow\left[\tan 1-\tan ^{-1} 1\right]=0\left\{\tan 1 \approx \tan \frac{\pi}{3}\right\} $
$\cot 1-\cot ^{-1} 1=\cot 1-\frac{\pi}{4} \approx 0.57-0.78<0 \text { and } \in(-1,0) \Rightarrow \left[\cot 1-\cot ^{-1} 1\right]=-1 $
$\sec 1-\sec ^{-1} 1=\sec 1-0< \sec \frac{\pi}{3}=2 . \text { As } \sec 1 \in(1,2) \Rightarrow \left[\sec 1-\sec ^{-1} 1\right]=1 $
$\text { and } \operatorname{cosec} 1-\operatorname{cosec}^{-1} 1=\operatorname{cosec} 1-\frac{\pi}{2}$
As $ \operatorname{cosec} \frac{\pi}{3}<\operatorname{cosec} 1<\operatorname{cosec} \frac{\pi}{4}<\frac{\pi}{2} \Rightarrow \operatorname{cosec} 1 \in\left(\frac{2}{\sqrt{3}}, \sqrt{2}\right)$
$\therefore \operatorname{cosec} 1-\frac{\pi}{2} \in(-1,0) \Rightarrow \left[\operatorname{cosec} 1-\operatorname{cosec}^{-1} 1\right]=-1$
Hence the value of given exprersion $=0-(-1)+0-(-1)+1-(-1)=1+1+1+1=4$