Question

The value of $10 \tan(\cot^{-1}3+\cot^{-1}7)$ is equal to

• A. $3$
• B. $5$
• C. $7$
• D. $9$
• E. $10$

Hint:

Formula Tip: When $AB < 1$, the formula $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$ works perfectly. If $AB > 1$, you must add $\pi$ to the result!

Answer 1

The Correct Option is B

Concept:
Inverse cotangent functions can be converted to inverse tangent functions using the property $\cot^{-1}x = \tan^{-1}\left(\frac{1}{x}\right)$ for $x > 0$. Then, the addition formula for inverse tangents $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$ is applied.

Step 1: Convert inverse cotangents to inverse tangents.
Using the reciprocal property: $$\cot^{-1}3 = \tan^{-1}\left(\frac{1}{3}\right)$$ $$\cot^{-1}7 = \tan^{-1}\left(\frac{1}{7}\right)$$ Substitute these into the expression: $$10 \tan\left( \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} \right)$$

Step 2: Apply the inverse tangent addition formula.
Use the formula $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A + B}{1 - AB}\right)$: $$\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1} \left( \frac{ \frac{1}{3} + \frac{1}{7} }{ 1 - (\frac{1}{3})(\frac{1}{7}) } \right)$$

Step 3: Simplify the numerator and denominator.
Numerator: $\frac{1}{3} + \frac{1}{7} = \frac{7 + 3}{21} = \frac{10}{21}$ Denominator: $1 - \left(\frac{1}{21}\right) = \frac{21 - 1}{21} = \frac{20}{21}$ Combine them into the inverse tangent: $$= \tan^{-1} \left( \frac{10/21}{20/21} \right)$$

Step 4: Calculate the inner value.
The denominators cancel out: $$= \tan^{-1} \left( \frac{10}{20} \right) = \tan^{-1}\left(\frac{1}{2}\right)$$ Our expression is now reduced to: $$10 \tan \left( \tan^{-1}\left(\frac{1}{2}\right) \right)$$

Step 5: Evaluate the final result.
Because tangent and inverse tangent are inverse functions of each other, $\tan(\tan^{-1}x) = x$: $$10 \left( \frac{1}{2} \right)$$ $$= 5$$ Hence the correct answer is (B) $5$.