CBSE Class 10 Mathematics Basic 2023: question 29(b)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. LHS $=\left(1+\frac{{\cos }^{2}A}{{\sin }^{2}A}\right)\left(1+\frac{{\sin }^{2}A}{{\cos }^{2}A}\right)$ [since $\frac{1}{\tan A}=\frac{\cos A}{\sin A}$, $\frac{1}{\cot A}=\frac{\sin A}{\cos A}$]

   1 mark

2. $=\left(\frac{{\sin }^{2}A+{\cos }^{2}A}{{\sin }^{2}A}\right)\left(\frac{{\cos }^{2}A+{\sin }^{2}A}{{\cos }^{2}A}\right)=\frac{1}{{\sin }^{2}A}\times \frac{1}{{\cos }^{2}A}$ [since ${\sin }^{2}A+{\cos }^{2}A=1$]

   1 mark

3. $=\frac{1}{{\sin }^{2}A(1-{\sin }^{2}A)}$ [since ${\cos }^{2}A=1-{\sin }^{2}A$]

   ½ mark

4. $=\frac{1}{{\sin }^{2}A-{\sin }^{4}A}=$ RHS. Hence proved.

   ½ mark

Answer: LHS = $\frac{1}{{\sin }^{2}A-{\sin }^{4}A}$ = RHS

Revise the chapter

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