Introduction to Trigonometry and Its Applications
NCERT Exemplar Ch 08
CBSE_NCERT_EXEMPLAR_CH08
Grade 10
Question:
If $\sin \theta + \cos \theta = \sqrt{2} \cos \theta$, then $\tan \theta$ is equal to:
$\sqrt{2} - 1$
$\sqrt{2} + 1$
$\dfrac{1}{\sqrt{2}}$
$\sqrt{2}$
Step-by-Step Solution
Key Concept: Divide both sides by $\cos \theta$.
Stepwise Solution:
Divide by $\cos \theta$: $\dfrac{\sin \theta}{\cos \theta} + 1 = \sqrt{2} \Rightarrow \tan \theta + 1 = \sqrt{2}$. [0.5 Mark]
$\tan \theta = \sqrt{2} - 1$. [0.5 Mark]
Marking Scheme:
• Dividing equation by $\cos \theta$: 0.5 Mark
• Solving for $\tan \theta = \sqrt{2} - 1$: 0.5 Mark
Correct Answer: $\sqrt{2} - 1$
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