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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
$\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}-\tan ^{-1} \frac{8}{19}=$
  • ✓
    $\frac{\pi}{4}$
  • B
    $\frac{\pi}{3}$
  • C
    $\frac{\pi}{6}$
  • D
    $\frac{2 \pi}{3}$
✓

Answer

Correct option: A.
$\frac{\pi}{4}$
(A) $\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}-\tan ^{-1} \frac{8}{19}$
$=\tan ^{-1}[\frac{\frac{3}{4}+\frac{3}{5}}{1-\frac{3}{4} \times \frac{3}{5}}]-\tan ^{-1} \frac{8}{19}$
$=\tan ^{-1} \frac{27}{11}-\tan ^{-1} \frac{8}{19}=\tan ^{-1}[\frac{\frac{27}{11}-\frac{8}{19}}{1+\frac{27}{11} \times \frac{8}{19}}]$
$=\tan ^{-1}(\frac{425}{425})=\tan ^{-1}(1)=\frac{\pi}{4}$
Alternate method:
$=\tan ^{-1}[\frac{\frac{3}{4}+\frac{3}{5}+\frac{8}{19}-\frac{3}{4} \times \frac{3}{5} \times \frac{8}{19}}{1-\frac{3}{4} \times \frac{3}{5}-\frac{3}{4} \times \frac{8}{19}-\frac{3}{5} \times \frac{8}{19}}]$ ...[Using Shortcut 4]
$=\tan ^{-1}(1)=\frac{\pi}{4}$

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