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

Concept: We solve each section of List-I step-by-step using inverse trigonometric identities.

Step 1: Solving Part (A).
\( Tan^{-1}x = Tan^{-1}8 - Tan^{-1}3 \). Using the identity \( Tan^{-1}a - Tan^{-1}b = Tan^{-1}\left(\frac{a-b}{1+ab}\right) \): \[ Tan^{-1}x = Tan^{-1}\left(\frac{8-3}{1+8\times3}\right) = Tan^{-1}\left(\frac{5}{25}\right) = Tan^{-1}\left(\frac{1}{5}\right) \implies x = \frac{1}{5} \quad \text{(Matches II)} \]

Step 2: Solving Part (B).
We know that \( Sin^{-1}x + Cos^{-1}x = \frac{\pi}{2} \implies Cos^{-1}x = \frac{\pi}{2} - Sin^{-1}x \). Substitute this into the equation: \[ Sin^{-1}x - \left(\frac{\pi}{2} - Sin^{-1}x\right) = \frac{\pi}{6} \implies 2Sin^{-1}x = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3} \] \[ Sin^{-1}x = \frac{\pi}{3} \implies x = sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \quad \text{(Matches III)} \]

Step 3: Solving Part (C).
Convert terms to tangent: Let \( sin^{-1}\frac{4}{5} = Tan^{-1}\frac{4}{3} \). Also, \( 2Tan^{-1}\frac{1}{3} = Tan^{-1}\left(\frac{2/3}{1-1/9}\right) = Tan^{-1}\left(\frac{2/3}{8/9}\right) = Tan^{-1}\frac{3}{4} \). \[ \text{Sum} = Tan^{-1}\frac{4}{3} + Tan^{-1}\frac{3}{4} = Tan^{-1}\frac{4}{3} + Cot^{-1}\frac{4}{3} = \frac{\pi}{2} \quad \text{(Matches IV)} \]

Step 4: Solving Part (D).
Right side: Let \( \theta = Tan^{-1}2 \implies tan\theta = 2 \implies sin\theta = \frac{2}{\sqrt{5}} \). Left side: Let \( Sec^{-1}\frac{1}{x} = \phi \implies sec\phi = \frac{1}{x} \implies tan\phi = \sqrt{\frac{1}{x^2}-1} \). Equating both sides: \[ \sqrt{\frac{1}{x^2}-1} = \frac{2}{\sqrt{5}} \implies \frac{1}{x^2} - 1 = \frac{4}{5} \implies \frac{1}{x^2} = \frac{9}{5} \implies x = \frac{\sqrt{5}}{3} \quad \text{(Matches I)} \] This full configuration cleanly corresponds to option (B).