CBSE Class 10 Mathematics Standard 2026: question 28

Hints

Open one hint at a time and try again before the next.

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 $=\frac{\frac{\sin \theta }{\cos \theta }}{1-\frac{\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{1-\frac{\sin \theta }{\cos \theta }}$

   ½ mark

2. $=\frac{{\sin }^2\theta }{\cos \theta (\sin \theta -\cos \theta )}+\frac{{\cos }^2\theta }{\sin \theta (\cos \theta -\sin \theta )}=\frac{{\sin }^2\theta }{\cos \theta (\sin \theta -\cos \theta )}-\frac{{\cos }^2\theta }{\sin \theta (\sin \theta -\cos \theta )}$

   ½ mark

3. $=\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta \cos \theta (\sin \theta -\cos \theta )}$

   ½ mark

4. Using $a^3-b^3=(a-b)(a^2+ab+b^2)$: $=\frac{(\sin \theta -\cos \theta )({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta )}{\sin \theta \cos \theta (\sin \theta -\cos \theta )}$

   ½ mark

5. $=\frac{{\sin }^2\theta }{\sin \theta \cos \theta }+\frac{{\cos }^2\theta }{\sin \theta \cos \theta }+\frac{\sin \theta \cos \theta }{\sin \theta \cos \theta }$

   ½ mark

6. $=\tan \theta +\cot \theta +1=$ RHS. Hence proved.

   ½ mark

Answer: LHS simplifies to $1+\tan \theta +\cot \theta$ = RHS. Hence proved.

Revise the chapter

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