coth^(-1)(2)+cosech^(-1)(-2√2)=

Correct answer: (A) log√32. Given, coth^(-1)(2)+cosech^(-1)(-2√2)=coth^(-1)(2)-cosech^(-1)(2√2) coth^(-1)2=1/2log(2+1)/(2-1)=1/2log3=log√3 (As coth^(-1)x=1/2log(x+1)/(x-1)) cosech^(-1)2√2=log(1+√(22^2+1))/(2√2)=log4/(2√2)=log√2 (As cosech^(-1)x=log(1+√(x^2+1))/x) ⇒coth^(-1)(2)-cosech^(-1)(2√2)=log√3-log√2=log√32

Q66 / 160 · TS EAMCET 2021 (04 Aug Shift 1) · Mathematics

$\cot h^{- 1} (2) + cosec h^{- 1} (- 2 \sqrt{2}) =$

Correct answer
A
$\log \sqrt{\frac{3}{2}}$
B
$\log \sqrt{6}$
C
$\log \frac{3}{\sqrt{2}}$
D
$\log \frac{3}{2}$

Solution

Given, $\cot h^{- 1} (2) + cosec h^{- 1} (- 2 \sqrt{2}) = \cot h^{- 1} (2) - cosec h^{- 1} (2 \sqrt{2})$

$\cot h^{- 1} (2) = \frac{1}{2} \log (\frac{2 + 1}{2 - 1}) = \frac{1}{2} \log (3) = \log \sqrt{3}$

(As $\cot h^{- 1} x = \frac{1}{2} \log (\frac{x + 1}{x - 1})$ )

$cosec h^{- 1} (2 \sqrt{2}) = \log (\frac{1 + \sqrt{{(2 \sqrt{2})}^{2} + 1}}{2 \sqrt{2}}) = \log (\frac{4}{2 \sqrt{2}}) = \log (\sqrt{2})$

(As $cosec h^{- 1} x = \log (\frac{1 + \sqrt{x^{2} + 1}}{x})$ )

$\Rightarrow \coth^{- 1} (2) - {cosech}^{- 1} (2 \sqrt{2}) = \log \sqrt{3} - \log \sqrt{2} = \log \sqrt{\frac{3}{2}}$

Hence, the correct answer is option: 'A'.

coth-1(2)+cosech-1(-22)=

Solution

Relevant Questions

$\cot h^{- 1} (2) + cosec h^{- 1} (- 2 \sqrt{2}) =$