Question

pls answer this question



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Answer 1

Given:

$\sf{\sqrt 3}{sin \theta = cos \theta}$

To Find:

$\sf{\dfrac{3cos^2 \theta + 2cos \theta}{3cos \theta +2}}$

Solution:

$\sf{\sqrt 3}{sin \theta = cos \theta}$

$\sf{\implies}{\sqrt 3 =}{\dfrac{cos \theta}{sin \theta}}$

$\sf{\implies}{\sqrt 3 =}{cot \theta}$

$\sf{\implies}{\dfrac{1}{\sqrt 3}}{=}{\dfrac{P}{B}}$

Comparing, we get, B= √3 and P=1

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From Pythagoras theorem,

$\sf{H=}{\sqrt{B^2 + P^2}}$

$\sf{H=}{\sqrt{{\sqrt 3}^2 + 1}}$

$\sf{H=}{\sqrt{4}}$

$\sf{H=2}$

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$\sf{cos \theta =}{\dfrac{\sqrt 3}{2}}$

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Substituting Values,

$\sf{\implies}{\dfrac{{\dfrac{9}{4}}+\sqrt 3}{{\dfrac{3\sqrt 3}{2}} + 2}}$

$\sf{\implies}{\dfrac{9+4\sqrt3}{6\sqrt 3 +8}}$

Rationalizing the denominator,

$\sf{implies}{\dfrac{54\sqrt 3 - 32\sqrt 3}{44}}$