CBSE Class 10 Mathematics Basic 2024: question 31

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 $=({\tan }^{2}A+{\sec }^{2}A+2\sec A\tan A)+({\tan }^{2}A+{\sec }^{2}A-2\sec A\tan A)=2({\tan }^{2}A+{\sec }^{2}A)$

   1 mark

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

   1½ marks

3. Using ${\cos }^{2}A=1-{\sin }^{2}A$: $=2\left(\frac{1+{\sin }^{2}A}{1-{\sin }^{2}A}\right)=$ RHS. Hence proved.

   ½ mark

Answer: LHS = 2(1 + sin²A)/(1 − sin²A) = RHS

Revise the chapter

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