CBSE Class 10 Mathematics Basic 2025: question 30

Hints

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Step-by-step solution

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1. LHS $=\frac{1+\frac{{\cos }^{2}A}{{\sin }^{2}A}}{1+\frac{{\sin }^{2}A}{{\cos }^{2}A}}=\frac{\frac{{\sin }^{2}A+{\cos }^{2}A}{{\sin }^{2}A}}{\frac{{\cos }^{2}A+{\sin }^{2}A}{{\cos }^{2}A}}=\frac{{\cos }^{2}A}{{\sin }^{2}A}={\cot }^{2}A$ [using ${\sin }^{2}A+{\cos }^{2}A=1$]

   1 mark

2. RHS: $\frac{1-\cot A}{1-\tan A}=\frac{\frac{\sin A-\cos A}{\sin A}}{\frac{\cos A-\sin A}{\cos A}}=\frac{\sin A-\cos A}{\sin A}\times \frac{\cos A}{-(\sin A-\cos A)}=-\frac{\cos A}{\sin A}=-\cot A$

   1½ marks

3. So RHS $=(-\cot A{)}^2={\cot }^{2}A=$ LHS. Hence proved.

   ½ mark

Answer: Both sides equal cot²A; hence proved

Revise the chapter

• Chapter: Introduction to Trigonometry (notes, formulas, where marks are lost and every past question)
• Topic: Proving Simple Trigonometric Identities