JEE Mains · Maths · STD 11 - Trigonometrical equations
A \(10\, inches\) long pencil \(\mathrm{AB}\) with mid point \(\mathrm{C}\) and a small eraser \(\mathrm{P}\) are placed on the horizontal top of a table such that \(\mathrm{PC}=\sqrt{5}\) inches and \(\angle \mathrm{PCB}=\tan ^{-1}(2)\). The acute angle through which the pencil must be rotated about \(\mathrm{C}\) so that the perpendicular distance between eraser and pencil becomes exactly \(1\, inch\) is:

- A \(\tan ^{-1}\left(\frac{3}{4}\right)\)
- B \(\tan ^{-1}(1)\)
- C \(\tan ^{-1}\left(\frac{4}{3}\right)\)
- D \(\tan ^{-1}\left(\frac{1}{2}\right)\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1}\left(\frac{3}{4}\right)\)
Step-by-step Solution
Detailed explanation
From figure, \(\sin \beta=\frac{1}{\sqrt{5}}\) \(\therefore \tan \beta=\frac{1}{2}\) \(\tan (\alpha+\beta)=2\) \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=2\) \(\frac{\tan \alpha+\frac{1}{2}}{1-\tan \alpha\left(\frac{1}{2}\right)}=2\)…
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