# Find the value of the given trigonometric equation.

$\dfrac{{\sin {{30}^0} + \tan {{45}^0} - \cos ec{{60}^0}}}{{\sec {{30}^0} + \cos {{60}^0} + \cot {{45}^0}}}$

Last updated date: 31st Mar 2023

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Answer

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Hint- To solve the above question we need only to put the values of the given terms from the trigonometry angles table. After substituting the values we need only to simplify the question.

Complete step-by-step solution -

Given equation is $\dfrac{{\sin {{30}^0} + \tan {{45}^0} - \cos ec{{60}^0}}}{{\sec {{30}^0} + \cos {{60}^0} + \cot {{45}^0}}}$

As we know that

$

\sin {30^0} = \dfrac{1}{2},\tan {45^0} = 1,{\text{cosec}}{60^0} = \dfrac{2}{{\sqrt 3 }}, \\

\sec {30^0} = \dfrac{2}{{\sqrt 3 }},\cos {60^0} = \dfrac{1}{2},\cot {45^0} = 1 \\

$

Substituting the above values in the given equation and simplifying further, we get

$

\Rightarrow \dfrac{{\sin {{30}^0} + \tan {{45}^0} - \cos ec{{60}^0}}}{{\sec {{30}^0} + \cos {{60}^0} + \cot {{45}^0}}} \\

\Rightarrow \dfrac{{\dfrac{1}{2} + 1 - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{2}{{\sqrt 3 }} + \dfrac{1}{2} + 1}} \\

\Rightarrow \dfrac{{\dfrac{{\sqrt 3 + 2\sqrt 3 - 4}}{{2\sqrt 3 }}}}{{\dfrac{{4 + \sqrt 3 + 2\sqrt 3 }}{{2\sqrt 3 }}}} \\

$

By simplifying further and cancelling the like term , we get

\[

\Rightarrow \dfrac{{\sqrt 3 + 2\sqrt 3 - 4}}{{4 + \sqrt 3 + 2\sqrt 3 }} \\

\Rightarrow \dfrac{{\sqrt 3 (1 + 2) - 4}}{{\sqrt 3 (1 + 2) + 4}} \\

\Rightarrow \dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}} \\

\]

Hence, the value of the given term is \[\dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}\]

Note- To solve these type of questions, you must remember the values of sine, cosine, tan, cot, cosec, cot from the trigonometry table for standard angles such as ${0^0},{30^0},{45^0},{60^0},{90^0}$ . The only thing left after this is to solve the equations using arithmetic operations such as addition, multiplication, division, subtraction, cross multiplication etc.

Complete step-by-step solution -

Given equation is $\dfrac{{\sin {{30}^0} + \tan {{45}^0} - \cos ec{{60}^0}}}{{\sec {{30}^0} + \cos {{60}^0} + \cot {{45}^0}}}$

As we know that

$

\sin {30^0} = \dfrac{1}{2},\tan {45^0} = 1,{\text{cosec}}{60^0} = \dfrac{2}{{\sqrt 3 }}, \\

\sec {30^0} = \dfrac{2}{{\sqrt 3 }},\cos {60^0} = \dfrac{1}{2},\cot {45^0} = 1 \\

$

Substituting the above values in the given equation and simplifying further, we get

$

\Rightarrow \dfrac{{\sin {{30}^0} + \tan {{45}^0} - \cos ec{{60}^0}}}{{\sec {{30}^0} + \cos {{60}^0} + \cot {{45}^0}}} \\

\Rightarrow \dfrac{{\dfrac{1}{2} + 1 - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{2}{{\sqrt 3 }} + \dfrac{1}{2} + 1}} \\

\Rightarrow \dfrac{{\dfrac{{\sqrt 3 + 2\sqrt 3 - 4}}{{2\sqrt 3 }}}}{{\dfrac{{4 + \sqrt 3 + 2\sqrt 3 }}{{2\sqrt 3 }}}} \\

$

By simplifying further and cancelling the like term , we get

\[

\Rightarrow \dfrac{{\sqrt 3 + 2\sqrt 3 - 4}}{{4 + \sqrt 3 + 2\sqrt 3 }} \\

\Rightarrow \dfrac{{\sqrt 3 (1 + 2) - 4}}{{\sqrt 3 (1 + 2) + 4}} \\

\Rightarrow \dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}} \\

\]

Hence, the value of the given term is \[\dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}\]

Note- To solve these type of questions, you must remember the values of sine, cosine, tan, cot, cosec, cot from the trigonometry table for standard angles such as ${0^0},{30^0},{45^0},{60^0},{90^0}$ . The only thing left after this is to solve the equations using arithmetic operations such as addition, multiplication, division, subtraction, cross multiplication etc.

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