CBSE Class 10 Mathematics Standard 2023: question 29(A)

Hints

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Step-by-step solution

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1. Take out the common factors:

   $\text{LHS}=\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-\cos A}=\frac{\sin A(1-2{\sin }^{2}A)}{\cos A(2{\cos }^{2}A-1)}$

   1 mark

2. Use ${\sin }^{2}A=1-{\cos }^{2}A$:

   $1-2{\sin }^{2}A=1-2(1-{\cos }^{2}A)=1-2+2{\cos }^{2}A=2{\cos }^{2}A-1$

   1 mark

3. $\text{LHS}=\frac{\sin A(2{\cos }^{2}A-1)}{\cos A(2{\cos }^{2}A-1)}=\frac{\sin A}{\cos A}=\tan A=\text{RHS}$

   Hence proved.

   1 mark

Answer: LHS simplifies to $\tan A$ = RHS (proved).

Revise the chapter

• Chapter: Introduction to Trigonometry (notes, formulas, where marks are lost and every past question)