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Introduction to Trigonometry
RD Sharma
CBSE
Grade 10

Question:

Evaluate: $\dfrac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$.

Step-by-Step Solution

Key Concept: Denominator $= 1$. Numerator $= 5(1/2)^2 + 4(2/\sqrt{3})^2 - 1^2 = 5/4 + 16/3 - 1 = \dfrac{15 + 64 - 12}{12} = \dfrac{67}{12}$.
Denominator $= \sin^2 30^\circ + \cos^2 30^\circ = 1$. [0.5 Mark]
Numerator $= 5\left(\dfrac{1}{4}\right) + 4\left(\dfrac{4}{3}\right) - 1 = \dfrac{5}{4} + \dfrac{16}{3} - 1$. [1.0 Mark]
$= \dfrac{15 + 64 - 12}{12} = \dfrac{67}{12}$. [0.5 Mark]

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🎯 Official CBSE Marking Scheme:
Evaluating denominator $= 1$: 0.5 Mark
Substituting values into numerator: 1.0 Mark
Evaluating fraction $= 67/12$: 0.5 Mark

Correct Answer:
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