If cos = 3 /5 , evaluate( sin−cot)/ 2tan – InterviewSolution

1.

If cos = 3 /5 , evaluate( sin−cot)/ 2tan

Answer»

$\huge\mathfrak\blue{Answer:}$

Given:

We have been given a VALUE of TRIGONOMETRIC function
$\sf{Cos\: x= \dfrac{3}{5}}$

To Find:

We have to find the value of
$\\ \sf{\left ( \dfrac{sin\: x - cot \: x}{2 tan \: x} \right ) }$

Information Used:

$\boxed{\sf{sin \: x = \left ( \dfrac{Perpendicular}{Hypotenuse} \right ) }} \\$
$\boxed{\sf{cos \: x = \left ( \dfrac{Base}{Hypotenuse} \right ) }} \\$
$\boxed{\sf{tan \: x = \left ( \dfrac{Perpendicular}{Base} \right ) }} \\$
$\boxed{\sf{cosec \: x = \left ( \dfrac{Base}{Perpendicular} \right ) }} \\$
$\boxed{\sf{sec \: x = \left ( \dfrac{Hypotenuse}{Base} \right ) }} \\$
$\boxed{\sf{cot \: x = \left ( \dfrac{Base}{Perpendicular} \right ) }} \\$

SOLUTION:

We have been given that :

$\boxed{\sf{\blue{Cos \: x = \dfrac{3}{5}} }}$

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$\bigstar \: \: \underline{\large\mathfrak\orange{We \: know \: that:}}$

$\hookrightarrow \sf{cos\: x = \dfrac{Base}{Hypotenuse}}$

$\hookrightarrow \sf{\dfrac{Base}{Hypotenuse } = \dfrac{3}{5}} \\ \\$

Let us ASSUME

$\sf{Base(B) = 3x}$

$\sf{Hypotenuse(H) = 5x}$

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$\bigstar \: \: \underline{\large\mathfrak\orange{Using \: Pythagoras \: Theorm} }$

$\implies \sf{ {(H)}^2 = {(P)}^2 + {(B)}^2}$

$\implies \sf{ {(5x)}^2 = P^2 + {(3x)}^2}$

$\implies \sf{ 25x^2 = P^2 = 9x^2}$

$\implies \sf{P^2 = 25x^2-9x^2}$

$\implies \sf{P^2 = 16x^2}$

Taking SQUARE root on Both sides

$\implies \sf{P = \sqrt{16x^2}}$

$\implies \sf{P = 4x}$

Hence Perpendicular of triangle = 4x

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Finding Values

$\implies \sf{sin \: x = \dfrac{P}{H}=\dfrac{4x}{5x} =\dfrac{4}{5} }$

$\implies \sf{cot \: x = \dfrac{B}{P}=\dfrac{3x}{4x}=\dfrac{3}{4}}$

$\implies \sf{tan \: x = \dfrac{P}{B}=\dfrac{4x}{3x}=\dfrac{4}{3}} \\$

Hence obtained values are

$\boxed{\sf{\green{sin \: x = \dfrac{4}{5} }}} \\$

$\boxed{\sf{\green{cot \: x = \dfrac{3}{4} }}} \\$

$\boxed{\sf{\green{tan \: x = \dfrac{4}{3} }}} \\$

______________________________

$\bigstar \: \: \underline{\large\mathfrak\orange{According \: to \: the \: Question:}}$

$\hookrightarrow \sf{\dfrac{sin \: x - cot \: x }{2tan \: x}}$

$\hookrightarrow \sf{\dfrac{\left ( \dfrac{4}{5} \right ) - \left ( \dfrac{3}{4} \right ) }{\left ( 2 \times \dfrac{4}{3} \right ) }}$

Solving the expression

$\hookrightarrow \sf{\dfrac{ \left ( \dfrac{16-15}{20} \right ) }{\left ( \dfrac{8}{3} \right ) }}$

Reciprocaling the terms

$\hookrightarrow \sf{\dfrac{1}{20} \times \dfrac{3}{8}}$

$\hookrightarrow \sf{\dfrac{3}{160}} \\$

$\boxed{\sf{\red{\dfrac{sin\: x - cot\: x }{2tan\: x} = \dfrac{3}{160}}}}$

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