Question

Match the items of List - I with those of List - II: \[ \begin{array}{ll} \text{List - I} & \text{List - II} \\ \hline \text{A) } Tan^{-1}3+Tan^{-1}x=Tan^{-1}8 \implies x= & \text{I) } \frac{\sqrt{5}}{3} \\ \text{B) } Sin^{-1}x-Cos^{-1}x=\frac{\pi}{6} \implies x= & \text{II) } \frac{1}{5} \\ \text{C) } sin^{-1}\frac{4}{5}+2~Tan^{-1}\frac{1}{3}= & \text{III) } \frac{\sqrt{3}}{2} \\ \text{D) } tan(Sec^{-1}\frac{1}{x})=sin(Tan^{-1}2), x>0 \implies x= & \text{IV) } \frac{\pi}{2} \\ & \text{V) } \frac{\pi}{3} \end{array} \] The correct match is:

• A. A-I, B-III, C-V, D-IV
• B. A-II, B-III, C-IV, D-I
• C. A-III, B-II, C-IV, D-V
• D. A-II, B-I, C-IV, D-V

Hint:

Recognizing complementary inputs can save you from a lot of calculations. In part C, notice that \( \frac{4}{3} \) and \( \frac{3}{4} \) are reciprocal arguments. This means \( Tan^{-1}\alpha + Tan^{-1}(1/\alpha) = \frac{\pi}{2} \) automatically for any positive \( \alpha \).

Answer 1

The Correct Option is B

Step 1: Solve part A.
From $\tan^{-1}3 + \tan^{-1}x = \tan^{-1}8$, write $\tan^{-1}x = \tan^{-1}8 - \tan^{-1}3 = \tan^{-1}\dfrac{8-3}{1+24} = \tan^{-1}\dfrac{1}{5}$. So $x = \tfrac{1}{5}$, matching II.
Step 2: Solve part B.
Use $\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}$. The equation becomes $2\sin^{-1}x = \tfrac{\pi}{6}+\tfrac{\pi}{2} = \tfrac{2\pi}{3}$, so $\sin^{-1}x = \tfrac{\pi}{3}$, giving $x = \tfrac{\sqrt3}{2}$, matching III.
Step 3: Solve part C.
Write $\sin^{-1}\tfrac{4}{5} = \tan^{-1}\tfrac{4}{3}$ and $2\tan^{-1}\tfrac{1}{3} = \tan^{-1}\tfrac{3}{4}$. Their sum is $\tan^{-1}\tfrac{4}{3} + \cot^{-1}\tfrac{4}{3} = \tfrac{\pi}{2}$, matching IV.
Step 4: Set up part D.
Right side: $\theta = \tan^{-1}2$ gives $\sin\theta = \tfrac{2}{\sqrt5}$. Left side: $\sec\phi = \tfrac{1}{x}$ gives $\tan\phi = \sqrt{\tfrac{1}{x^2}-1}$.
Step 5: Solve part D.
Set $\sqrt{\tfrac{1}{x^2}-1} = \tfrac{2}{\sqrt5}$. Square: $\tfrac{1}{x^2}-1 = \tfrac{4}{5}$, so $\tfrac{1}{x^2} = \tfrac{9}{5}$ and $x = \tfrac{\sqrt5}{3}$, matching I.
Step 6: Combine the matches.
A-II, B-III, C-IV, D-I. \[ \boxed{\text{A-II, B-III, C-IV, D-I}} \]