Introduction to Trigonometry
CBSE 2026 Board Exam Set 1 (Code 30/7/1)
CBSE_BOARD_PYQ_2026_30_7_1
Grade 10
Question:
[Section C]
Prove that: $\dfrac{\sin A - \cos A + 1}{\sin A + \cos A - 1} = \sec A + \tan A$.
OR
If $\sin(A + 2B) = \dfrac{\sqrt{3}}{2}$ and $\cos(A + 4B) = 0$, where $0^\circ < A + 4B \le 90^\circ$, find $A$ and $B$.
Step-by-Step Solution
Key Concept: Main: Divide numerator and denominator by $\cos A$. OR: Linear equations in $A$ and $B$.
[Main Question Solution]
Divide by $\cos A$: $\dfrac{\tan A - 1 + \sec A}{\tan A + 1 - \sec A}$. Using $1 = \sec^2 A - \tan^2 A$. [1.5 Marks]
Factoring and simplifying yields $\sec A + \tan A$. Verified. [1.5 Marks]
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[OR Choice Question Solution]
$\sin(A + 2B) = \sqrt{3}/2 = \sin 60^\circ \Rightarrow A + 2B = 60^\circ$. Eq (1). [1.0 Mark]
$\cos(A + 4B) = 0 = \cos 90^\circ \Rightarrow A + 4B = 90^\circ$. Eq (2). [1.0 Mark]
Subtracting (1) from (2): $2B = 30^\circ \Rightarrow B = 15^\circ$. $A = 30^\circ$. [1.0 Mark]
Correct Answer: Main: Proof shown | OR: A = 30°, B = 15°
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