JEE Advanced 2024 Paper 2 Q01 Mathematics Trigonometry Inverse Trigonometric Functions Medium

JEE Advanced 2024 Paper 2 · Q01 · Inverse Trigonometric Functions

Considering only the principal values of the inverse trigonometric functions, the value of $$\tan\left(\sin^{-1}\!\left(\frac{3}{5}\right) - 2\cos^{-1}\!\left(\frac{2}{\sqrt{5}}\right)\right)$$ is

  1. A. $\dfrac{7}{24}$
  2. B. $-\dfrac{7}{24}$
  3. C. $-\dfrac{5}{24}$
  4. D. $\dfrac{5}{24}$
Reveal answer + step-by-step solution

Correct answer:B

Solution

With principal values, $\sin^{-1}(3/5) = \tan^{-1}(3/4)$ and $\cos^{-1}(2/\sqrt{5}) = \tan^{-1}(1/2)$. Using $2\tan^{-1}(1/2) = \tan^{-1}\!\left(\dfrac{2 \cdot 1/2}{1 - 1/4}\right) = \tan^{-1}(4/3)$. Hence the expression equals $\tan\!\left(\tan^{-1}(3/4) - \tan^{-1}(4/3)\right) = \dfrac{3/4 - 4/3}{1 + (3/4)(4/3)} = \dfrac{(9-16)/12}{2} = -\dfrac{7}{24}$. Answer: (B).

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