If \(a + \dfrac{\pi}{2} < 2 \tan^{-1} x + 3 \cot^{-1} x < b\) then 'a' and 'b' are respectively.

Question

Question

Answer 1

Concept:

\(\rm \tan^{-1}x+\cot^{-1}x=\frac{π}{2}\)

0 < cot^-1 x < π

Calculation:

0 < cot^-1 x < π

\( \rm ⇒ π+0<2\left(\frac{π}{2}\right)+\cot ^{-1} x<π+π \)

\(\because \tan ^{-1} x+\cot ^{-1} x=\frac{π}{2}\)

\(\rm ⇒ π<2\left(\tan ^{-1} x+\cot ^{-1} x\right)+\cot ^{-1} x<2 π\)

\(\rm ⇒ π<2 \tan ^{-1} x+3 \cot ^{-1} x<2 π \) ---(1)

After comparison,

\( a+\frac{π}{2}\) < 2 tan^-1 x + 3 cot^-1 x < b

⇒ \(\rm a=\frac{π}{2} \)

\(\rm b=2 π \)

Hence, option (2) is correct.