CBSE Class 12 Mathematics sample paper 2025-26: question 21(A)

Hints

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Step-by-step solution

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1. The principal value branch of ${\tan }^{-1}$ is $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, so ${\tan }^{-1}(-1)=-\frac{\pi }{4}$. The expression becomes $\tan \left(-\frac{\pi }{4}+\frac{\pi }{3}\right)=\tan \left(\frac{\pi }{3}-\frac{\pi }{4}\right)$.

   ½ mark

2. Using $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$: $\tan \left(\frac{\pi }{3}-\frac{\pi }{4}\right)=\frac{\tan \frac{\pi }{3}-\tan \frac{\pi }{4}}{1+\tan \frac{\pi }{3}\tan \frac{\pi }{4}}=\frac{\sqrt{3}-1}{1+\sqrt{3}}$.

   1 mark

3. Rationalising: $\frac{(\sqrt{3}-1{)}^2}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$.

   ½ mark

Answer: $\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$

Revise the chapter

• Chapter: Inverse Trigonometric Functions (notes, formulas, where marks are lost and every past question)