CBSE Class 10 Mathematics Standard 2024: question 29

Hints

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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 $=\frac{\frac{\sin A}{\cos A}}{\frac{\sin A-\cos A}{\sin A}}+\frac{\frac{\cos A}{\sin A}}{\frac{\cos A-\sin A}{\cos A}}$

   ½ mark

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

   1 mark

3. Using $a^3-b^3=(a-b)(a^2+b^2+ab)$: $=\frac{(\sin A-\cos A)({\sin }^{2}A+{\cos }^{2}A+\sin A\cos A)}{\sin A\cos A(\sin A-\cos A)}=\frac{1+\sin A\cos A}{\sin A\cos A}$ [since ${\sin }^{2}A+{\cos }^{2}A=1$]

   1 mark

4. $=\frac{1}{\sin A\cos A}+1=1+\sec A\csc A=$ RHS. Hence proved.

   ½ mark

Answer: LHS simplifies to $1+\sec A\csc A$ = RHS; hence proved.

Revise the chapter

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