MCQ
$\cos \left[\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{2}\right)\right]= ........ $
- A$\frac{\sqrt{3}}{2}$
- B$\frac{1}{2}$
- ✓$\frac{1}{\sqrt{2}}$
- D$\frac{\pi}{4}$
✓
Answer
Correct option: C.
$\frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{2}}$
$\cos \left[\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)\right]$
$=\cos \left[\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3} \times \frac{1}{2}}\right)\right]$
$[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)],$
if $[x>0, y>0, x y<1]$
$\therefore=\cos \left[\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)\right]$
$=\cos \left[\tan ^{-1}(1)\right]=\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$
$\cos \left[\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)\right]$
$=\cos \left[\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3} \times \frac{1}{2}}\right)\right]$
$[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)],$
if $[x>0, y>0, x y<1]$
$\therefore=\cos \left[\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)\right]$
$=\cos \left[\tan ^{-1}(1)\right]=\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$
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