Introduction to Trigonometry
CBSE 2026 Board Exam Set 1 (Code 30/7/1)
CBSE_BOARD_PYQ_2026_30_7_1
Grade 10
Question:
[Section B]
If $\tan A = \sqrt{2} - 1$, prove that $\dfrac{\tan A}{1 + \tan^2 A} = \dfrac{1}{2\sqrt{2}}$.
OR
If $\operatorname{cosec} \theta = \dfrac{13}{12}$, find the value of $\tan \theta + \cot \theta$.
Step-by-Step Solution
Key Concept: Main: Direct substitution and rationalization. OR: $\sin \theta = 12/13 \Rightarrow \cos \theta = 5/13$.
[Main Question Solution]
$\text{LHS} = \dfrac{\sqrt{2}-1}{1 + (\sqrt{2}-1)^2} = \dfrac{\sqrt{2}-1}{1 + 2 - 2\sqrt{2} + 1} = \dfrac{\sqrt{2}-1}{4 - 2\sqrt{2}}$. [1.0 Mark]
$= \dfrac{\sqrt{2}-1}{2\sqrt{2}(\sqrt{2}-1)} = \dfrac{1}{2\sqrt{2}} = \text{RHS}$. Hence proved. [1.0 Mark]
-------------------------------
[OR Choice Question Solution]
$\operatorname{cosec} \theta = 13/12 \Rightarrow \cot \theta = 5/12$ and $\tan \theta = 12/5$. [1.0 Mark]
$\tan \theta + \cot \theta = \dfrac{12}{5} + \dfrac{5}{12} = \dfrac{144 + 25}{60} = \dfrac{169}{60}$. [1.0 Mark]
Correct Answer: Main: Proof shown | OR: 169/60
Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.