Evaluate: (5cos^2 60 + 4 sec ^ 2 30 - tan ^2 45) / (sin^2 30

Q) Evaluate: $\frac{(5 \cos ^2 60 ^\circ + 4 \sec ^2 30 ^\circ - \tan ^2 45 ^\circ)}{(\sin ^2 30 ^\circ + \cos ^2 30 ^\circ)}$

(Q 22 A – 30/2/2 – CBSE 2026 Question Paper)

Ans:

Let’s solve numerator and denominator one by one and then combine:

Step 1: Solving for numerator:

(5 cos ² 60⁰ + 4 sec ² 30⁰ – tan ² 45⁰)

= 5 $(\frac{1}{2})^2 + 4 (\frac{1}{2})^2 - (1)^2$ (by putting values)

= 5 $(\frac{1}{4}) + 4(\frac{1}{4}) - 1 = (\frac{5}{4}) + (\frac{4}{4}) - 1$

= $\frac{9}{4} - 1 = \frac{(9 - 4)}{4}$

= $\frac{5}{4}$

Step 2: Solving for denominator:

(sin ² 30⁰ + cos ² 30⁰)

= $(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$ (by putting values)

= $\frac{1}{4} + \frac{3}{4}$

= $\frac{4}{4} = 1$

Step 3: Let’s substitute values from step 1 and step 2:

$\frac{(5 \cos ^2 60 ^\circ + 4 \sec ^2 30 ^\circ - \tan ^2 45 ^\circ)}{(\sin ^2 30 ^\circ + \cos ^2 30 ^\circ)}$

= $\frac{\frac{5}{4}}{1}$

= $\frac{5}{4}$

Therefore, the value of given expression is $\frac{5}{4}$ .

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