By signing in you confirm you are 18 or older, or have your parent’s or guardian’s consent.

NDA 1 2024 Maths, question 44

Topic: Trigonometry

Q44

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

  1. (a) −π4-\frac{\pi}{4}
  2. (b) π4\frac{\pi}{4}Correct answer
  3. (c) tan⁡−1(a2−b2a2+b2)\tan ^{-1}\left( \frac{a^{2}-b^{2}}{a^{2}+b^{2}} \right)
  4. (d) tan⁡−1(2aba2+b2)\tan ^{-1}\left( \frac{2ab}{a^{2}+b^{2}} \right)

Solution.

Answer: (b) π4\frac{\pi}{4}

  1. Let t=abt = \frac{a}{b}. Then a−ba+b=t−1t+1\frac{a - b}{a + b} = \frac{t - 1}{t + 1}.
  2. tan⁡(A−B)=t−t−1t+11+t⋅t−1t+1=t2+1t2+1=1\tan(A - B) = \frac{t - \frac{t - 1}{t + 1}}{1 + t\cdot\frac{t - 1}{t + 1}} = \frac{t^{2} + 1}{t^{2} + 1} = 1.
  3. So the difference is π4\frac{\pi}{4}.

Trap: This holds when ab>−1\frac{a}{b} > -1; for ab<−1\frac{a}{b} < -1 the principal values give −3π4-\frac{3\pi}{4}, which is not an option.