The value of \(cot^{-1}\left(cot\frac{5\pi}{4}\right)\)isA. \(\frac{\

1.	The value of \(cot^{-1}\left(cot\frac{5\pi}{4}\right)\)isA. \(\frac{\pi}{4}\)B. \(\frac{-\pi}{4}\)C. \(\frac{3\pi}{4}\)D. none of these
Answer» Correct Answer is \(\frac{\pi}{4}\) Now, let x = \(cot^{-1}\left(cot\frac{5\pi}{4}\right)\) ⇒ cot x =cot ( \(\frac{5\pi}{4}\)) Here range of principle value of cot is [ \(-\frac{\pi}{2},\frac{\pi}{2}\)] ⇒ x = \(\frac{5\pi}{4}\)∉ [ \(-\frac{\pi}{2},\frac{\pi}{2}\)] Hence for all values of x in range [ \(-\frac{\pi}{2},\frac{\pi}{2}\)] ,the value of \(cot^{-1}\left(cot\frac{5\pi}{4}\right)\) is ⇒ cot x =cot (π+ \(\frac{\pi}{4}\)) (\(\because\) cot ( \(\frac{5\pi}{4}\))= cot ( π+ \(\frac{\pi}{4}\) ) ) ⇒ cot x =cot ( \(\frac{\pi}{4}\)) ( \(\because\)cot (π + θ)= cot θ) ⇒ x =\(\frac{\pi}{4}\)

The value of \(cot^{-1}\left(cot\frac{5\pi}{4}\right)\)isA. \(\frac{\pi}{4}\)B. \(\frac{-\pi}{4}\)C. \(\frac{3\pi}{4}\)D. none of these

Answer»

Correct Answer is \(\frac{\pi}{4}\)

Now, let x = \(cot^{-1}\left(cot\frac{5\pi}{4}\right)\)

⇒ cot x =cot ( \(\frac{5\pi}{4}\))

Here range of principle value of cot is [ \(-\frac{\pi}{2},\frac{\pi}{2}\)]

⇒ x = \(\frac{5\pi}{4}\)∉ [ \(-\frac{\pi}{2},\frac{\pi}{2}\)]

Hence for all values of x in range [ \(-\frac{\pi}{2},\frac{\pi}{2}\)] ,the value of

\(cot^{-1}\left(cot\frac{5\pi}{4}\right)\) is

⇒ cot x =cot (π+ \(\frac{\pi}{4}\)) (\(\because\) cot ( \(\frac{5\pi}{4}\))= cot ( π+ \(\frac{\pi}{4}\) ) )

⇒ cot x =cot ( \(\frac{\pi}{4}\)) ( \(\because\)cot (π + θ)= cot θ)

⇒ x =\(\frac{\pi}{4}\)