ExamBro
ExamBro
enEnglishhiहिन्दीguગુજરાતી
JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

\(\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)\) ની કિમંત મેળવો.

  1. A \(1\)
  2. B \(2\)
  3. C \(\frac{1}{4}\)
  4. D \(\frac{5}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2\)

Step-by-step Solution

Detailed explanation

\(\tan \left(2\left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)\right)\) \(=\tan \left[2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{2}\right)\right]\) \(=2\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app