NDA 2026 — Mathematics PYQ

To solve this, we use the formula for $2{\tan }^{-1}(x)$ and the tangent subtraction formula $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$.

Step 1: Simplify $2{\tan }^{-1}\left(\frac{1}{2}\right)$

Using the identity $2{\tan }^{-1}(x)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$:

$$2{\tan }^{-1}\left(\frac{1}{2}\right)={\tan }^{-1}\left(\frac{2\cdot \frac{1}{2}}{1-(\frac{1}{2}{)}^2}\right)={\tan }^{-1}\left(\frac{1}{1-\frac{1}{4}}\right)={\tan }^{-1}\left(\frac{1}{\frac{3}{4}}\right)={\tan }^{-1}\left(\frac{4}{3}\right)$$

Step 2: Substitute back into the original expression

The expression becomes:

$$\tan \left[{\tan }^{-1}\left(\frac{4}{3}\right)-\frac{\pi }{4}\right]$$

Step 3: Apply the tangent subtraction formula

Let $A={\tan }^{-1}\left(\frac{4}{3}\right)$ and $B=\frac{\pi }{4}$. Then $\tan A=\frac{4}{3}$ and $\tan B=\tan \left(\frac{\pi }{4}\right)=1$.

$$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$

$$\tan \left[{\tan }^{-1}\left(\frac{4}{3}\right)-\frac{\pi }{4}\right]=\frac{\frac{4}{3}-1}{1+(\frac{4}{3})(1)}$$

Step 4: Final Calculation

$$\frac{\frac{4}{3}-\frac{3}{3}}{1+\frac{4}{3}}=\frac{\frac{1}{3}}{\frac{7}{3}}=\frac{1}{7}$$