Question

Evaluate \[ \tan^{-1}\left(\frac12\right) +\tan^{-1}\left(\frac18\right) +\tan^{-1}\left(\frac1{18}\right) +\tan^{-1}\left(\frac1{32}\right). \]

• A. \[ \tan^{-1}\left(\frac35\right) \]
• B. \[ \tan^{-1}\left(\frac58\right) \]
• C. \[ \tan^{-1}\left(\frac34\right) \]
• D. \[ \tan^{-1}\left(\frac45\right) \] \bigskip

Hint:

For sums of inverse tangents, repeatedly use \[ \tan^{-1}a+\tan^{-1}b = \tan^{-1} \left( \frac{a+b}{1-ab} \right), \] whenever \(ab<1\). Pairing terms cleverly often leads to a simple result.

Answer 1

The Correct Option is D

Concept: Use the identity \[ \tan^{-1}a-\tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right), \] whenever \(ab>-1\). A useful observation is \[ \tan^{-1}\left(\frac1n\right) = \tan^{-1}(n+1)-\tan^{-1}(n), \] provided \[ \frac{(n+1)-n}{1+n(n+1)} = \frac1{n^2+n+1}. \] This creates a telescoping sum.

Step 1: Express each term as a difference of inverse tangents. Since \[ \frac12=\frac1{1^2+1+1}, \] we have \[ \tan^{-1}\left(\frac12\right) = \tan^{-1}(2)-\tan^{-1}(1). \] Similarly, \[ \frac18=\frac1{2^2+2+2}, \] hence \[ \tan^{-1}\left(\frac18\right) = \tan^{-1}(3)-\tan^{-1}(2). \] Also, \[ \frac1{18}=\frac1{3^2+3+6}, \] and \[ \tan^{-1}\left(\frac1{18}\right) = \tan^{-1}(4)-\tan^{-1}(3). \] Finally, \[ \tan^{-1}\left(\frac1{32}\right) = \tan^{-1}(5)-\tan^{-1}(4). \]

Step 2: Add all four expressions. Therefore, \[ S= \left[\tan^{-1}(2)-\tan^{-1}(1)\right] +\left[\tan^{-1}(3)-\tan^{-1}(2)\right] \] \[ +\left[\tan^{-1}(4)-\tan^{-1}(3)\right] +\left[\tan^{-1}(5)-\tan^{-1}(4)\right]. \] All intermediate terms cancel: \[ S = \tan^{-1}(5)-\tan^{-1}(1). \]

Step 3: Use the subtraction formula. \[ S = \tan^{-1} \left( \frac{5-1}{1+5} \right). \] \[ = \tan^{-1} \left( \frac46 \right). \] \[ = \tan^{-1} \left( \frac23 \right). \] This does not match any option, so let us verify the third and fourth terms correctly. Observe that \[ \frac1{18} = \frac{4-3}{1+12}, \] hence \[ \tan^{-1}\left(\frac1{13}\right) = \tan^{-1}(4)-\tan^{-1}(3), \] not \(\tan^{-1}\left(\frac1{18}\right)\). Instead, \[ \frac1{18} = \frac{6-3}{1+18} = \frac3{19}, \] which is not directly telescoping. Therefore, use pairwise addition.

Step 4: Add the first two terms. \[ \tan^{-1}\left(\frac12\right) +\tan^{-1}\left(\frac18\right) = \tan^{-1} \left( \frac{\frac12+\frac18} {1-\frac1{16}} \right). \] \[ = \tan^{-1} \left( \frac{5/8}{15/16} \right) = \tan^{-1}\left(\frac23\right). \]

Step 5: Add the next two terms. \[ \tan^{-1}\left(\frac1{18}\right) +\tan^{-1}\left(\frac1{32}\right) = \tan^{-1} \left( \frac{\frac1{18}+\frac1{32}} {1-\frac1{576}} \right). \] \[ = \tan^{-1} \left( \frac{25/288}{575/576} \right) = \tan^{-1}\left(\frac{2}{23}\right). \] Hence, \[ S = \tan^{-1}\left(\frac23\right) + \tan^{-1}\left(\frac2{23}\right). \] Again applying the addition formula, \[ S = \tan^{-1} \left( \frac{\frac23+\frac2{23}} {1-\frac4{69}} \right). \] \[ = \tan^{-1} \left( \frac{52/69}{65/69} \right). \] \[ = \tan^{-1}\left(\frac45\right). \]

Step 6: Write the final answer. \[ \boxed{\tan^{-1}\left(\frac45\right)} \]