CBSE Class 10 Mathematics Standard 2026: question 26(b)

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

   ½ mark

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

   1 mark

3. Simplifying each bracket:
   $=1+\frac{{\cos }^{2}A+{\sin }^{2}A}{{\sin }^{2}A{\cos }^{2}A}+2\cdot \frac{{\sin }^{2}A+{\cos }^{2}A}{\sin A\cos A}$
   $=1+\frac{1}{{\sin }^{2}A{\cos }^{2}A}+\frac{2}{\sin A\cos A}$
   $=1+{\sec }^{2}A{\text{cosec}}^{2}A+2\sec A\text{cosec}A$

   1 mark

4. This has the form $1+2t+t^2$ with $t=\sec A\text{cosec}A$:
   $=(1+\sec A\text{cosec}A{)}^2=$ RHS.
   Hence proved.

   ½ mark

Answer: LHS simplifies to $1+2\sec A\text{cosec}A+{\sec }^{2}A{\text{cosec}}^{2}A=(1+\sec A\text{cosec}A{)}^2$ = RHS.

Revise the chapter

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