Question

give me a answer of
$if \: \: \tan( \alpha ) = \sqrt{3} then \: find \: the \: value \: of \: 2 \sin( \alpha ) + \cos( \alpha )$
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Answer 1

Trigonometry

The following are the tips and concept that can be use to find the solution:

• Having a basic knowledge of Trigonometric ratios.

• Trigonometric ratios are sin, cos, tan, cot, sec, cosec.

• Relationship between sides and T ratios

• Pythagoras theorem: The Pythagoras theorem states that if a triangle is right-angled, then the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Relationship between sides and T ratios:

• sin(θ) = Perpendicular/Hypotenuse
• cos(θ) = Base/Hypotenuse
• tan(θ) = Perpendicular/Base
• cot(θ) = Base/Perpendicular
• sec(θ) = Hypotenuse/Base
• cosec(θ) = Hypotenuse/Perpendicular

We have been given that,

$\longrightarrow \tan(\alpha) = \sqrt{3}$

With this information, we've been asked to calculate the value of $2\sin(\alpha) + \cos(\alpha)$.

We know that,

$\tan(\alpha) = \dfrac{P}{B} = \dfrac{\sqrt{3}}{1}$

By using Pythagoras theorem and substituting the known values in it, we obtain the following results:

$\implies H^2 = P^2 + B^2 \\ \\ \implies H^2 = (\sqrt{3})^2 + (1)^2 \\ \\ \implies H^2 = 3 + 1 \\ \\ \implies H^2 = 4 \\ \\ \implies H = \sqrt{4} \\ \\ \implies H = 2$

Now, we know that,

$\sin(\alpha) = \dfrac{P}{H} = \dfrac{\sqrt{3}}{2}$

$\cos(\alpha) = \dfrac{B}{H} = \dfrac{1}{2}$

Now, consider,

$\implies 2\sin(\alpha) + \cos(\alpha)$

By substituting the known values in the above equation, we get the following results:

$\implies 2 \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \\ \\ \implies \dfrac{2\sqrt{3}}{2} + \dfrac{1}{2} \\ \\ \implies \boxed{\dfrac{2\sqrt{3} + 1}{2}}$

Hence, this is our expected solution.