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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2024 · Mathematics · Inverse Trigonometric Functions

What is ^ -1 ((a)/(b))- ^ -1 ((a-b)/(a+b))equal to?

What is tan⁡−1(ab)−tan⁡−1(a−ba+b)equalto?\tan ^{-1} \left(\frac{a}{b}\right)-\tan ^{-1} \left(\frac{a-b}{a+b}\right)equal to?

Options

  1. A−π4-\frac{\pi}{4}
  2. Bπ4\frac{\pi}{4}
  3. Ctan⁡−1(a2−b2a2+b2)\tan ^{-1} \left(\frac{a^2-b²}{a^2+b²}\right)
  4. Dtan⁡−1(2aba2+b2)\tan ^{-1} \left(\frac{2 a b}{a^2+b²}\right)

Correct answer

B. π4\frac{\pi}{4}

Step-by-step solution

tan⁡−1(ab)−tan⁡−1(a−ba+b)=tan⁡−1(ab)−tan⁡−1(ab−11+ab⋅1)=tan⁡−1ab−tan⁡−1ab+tan⁡−11=tan⁡−1=tan⁡−1×tan⁡45∘=45∘=π4.\begin{aligned} & \tan ^{-1}\left(\frac{a}{b}\right)-\tan ^{-1}\left(\frac{a-b}{a+b}\right) \\ &=\tan ^{-1}\left(\frac{a}{b}\right)-\tan ^{-1}\left(\frac{\frac{a}{b}-1}{1+\frac{a}{b} \cdot 1}\right) \\ &=\tan ^{-1} \frac{a}{b}-\tan ^{-1} \frac{a}{b}+\tan ^{-1} 1 \\ &=\tan ^{-1}=\tan ^{-1} \times \tan 45^{{}^{\circ}} \\ &=45^{{}^{\circ}}=\frac{\pi}{4}.\end{aligned}
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