Question

$solve \: it...$

$\frac{5 \: { \cos }^{2 \:} \: 60 \: + \: 4 \: \: { \sec }^{2} \: \: 30 \: - \: { \tan }^{2} \: 45}{ { \sin }^{2} \: 30 \: + \: { \cos }^{2} \: 30 }$
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Answer 1

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{Question}}}}}}}$

$solve \: it...$

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

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{Solution}}}}}}}$

$\sf\implies \dfrac{5 \: { \cos }^{2 \:} \: 60 \: + \: 4 \: \: { \sec }^{2} \: \: 30 \: - \: { \tan }^{2} \: 45}{ { \sin }^{2} \: 30 \: + \: { \cos }^{2} \: 30 }$

$\sf{\bold{\blue{\boxed{\boxed{sin^2\theta + cos^2 \theta = 1 }}}}}$

$\sf\implies\dfrac{ 5 cos^2 60 + 4 sec^2 30 - tan^2 45 }{1}$

$\sf{\bold{\blue{\boxed{\boxed{Cos 60 = \dfrac{1}{2}}}}}}$

$\sf{\bold{\blue{\boxed{\boxed{sec 30 = \dfrac{2}{\sqrt 3}}}}}}$

$\sf{\bold{\blue{\boxed{\boxed{tan 45 = 1}}}}}$

• putting triangular values

$\sf\implies 5\times (\dfrac{1}{2})^2 + 4\times (\dfrac{2}{\sqrt 3})^2 - 1$

$\sf\implies 5 \times(\dfrac{1}{4}) + 4\times (\dfrac{4}{3}) - 1$

$\sf\implies \dfrac{5}{4} + \dfrac{16}{3} - 1$

$\sf\implies\dfrac{15 + 64 -12 }{12}$

$\sf\implies \dfrac{67}{12}$

$\sf{\bold{\green{\boxed{\boxed{\dfrac{67}{12}}}}}}$

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Answer 2

Answer :

Given -

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

To Find -

• Value ?

Solution -

To solve such questions, we need to learn the values of trigonometric ratios of 0°, 30°, 45°, 60° and 90°.

Also, one Identity will be used : sin²∅ + cos²∅ = 1.

Here, we need the following values :

• cos 60 = ½
• sec 30 = 2/√3
• tan 45 = 1

Now, solving the question :

Substituting the values -

$\sf{=\dfrac{5\:\times{(1/2)}^{2}+ 4\:\times{(2/\sqrt{3})}^{2} -{1}^{2}}{1}}$

$\sf{= 5\:\times 1/4+ 4\:\times 4/3 - 1}$

$\sf{=\dfrac{5}{4}+ \dfrac{16}{3} - 1}$

Taking LCM -

$\sf{=\dfrac{15+64-12}{12}}$

$\sf{=\dfrac{79-12}{12}}$

$\sf{=\dfrac{67}{12}}$

Hence, the value = 67/12.