CBSE Class 10 Mathematics Basic 2025: question 28

Hints

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Step-by-step solution

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1. LHS $={\csc }^{2}A+{\sin }^{2}A+2\csc A\sin A+{\sec }^{2}A+{\cos }^{2}A+2\sec A\cos A$

   1 mark

2. Since $\csc A\sin A=1$ and $\sec A\cos A=1$, and using ${\csc }^{2}A=1+{\cot }^{2}A$, ${\sec }^{2}A=1+{\tan }^{2}A$:
   LHS $=({\sin }^{2}A+{\cos }^{2}A)+(1+{\cot }^{2}A)+(1+{\tan }^{2}A)+2+2$

   1 mark

3. $=1+1+1+4+{\tan }^{2}A+{\cot }^{2}A=7+{\tan }^{2}A+{\cot }^{2}A=$ RHS. Hence proved.

   1 mark

Answer: LHS simplifies to $7+{\tan }^{2}A+{\cot }^{2}A$ = RHS; proved.

Revise the chapter

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