Quadratic Equations
CBSE 2026 Board Exam Set 3 (Code 30/1/3)
CBSE_BOARD_PYQ_2026_30_1_3
Grade 10
Question:
[Section C]
Solve for $x$: $9x^2 - 9(a+b)x + (2a^2 + 5ab + 2b^2) = 0$.
OR
Prove that: $\dfrac{\tan A}{1 + \sec A} - \dfrac{\tan A}{1 - \sec A} = 2 \text{cosec } A$.
Step-by-Step Solution
Key Concept: Factorise quadratic equation OR Convert trigonometric expressions to sin and cos.
Main: $2a^2 + 5ab + 2b^2 = (2a+b)(a+2b)$. Middle term split: $-[3(2a+b) + 3(a+2b)] = -9(a+b)$. Roots are $x = \dfrac{2a+b}{3}$ and $x = \dfrac{a+2b}{3}$. [3.0 Marks]
OR: $\text{LHS} = \tan A \left[ \dfrac{(1 - \sec A) - (1 + \sec A)}{1 - \sec^2 A} \right] = \tan A \left[ \dfrac{-2\sec A}{-\tan^2 A} \right] = \dfrac{2\sec A}{\tan A} = \dfrac{2/\cos A}{\sin A/\cos A} = \dfrac{2}{\sin A} = 2\text{cosec } A = \text{RHS}$. [3.0 Marks]
Correct Answer: x = (2a+b)/3 or x = (a+2b)/3 OR Proof completed
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