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# Solve the equation and find its value: $\dfrac{{{\text{sin30 + tan45 - cosec60}}}}{{{\text{sec30 + cos60 + cot45}}}}$.

Last updated date: 23rd Mar 2023
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Hint: In order to solve this question, you have to put the value of the trigonometric ratios for the particular angles provided in the question and simplify it to get the correct answer.

The given equation is $\dfrac{{{\text{sin30 + tan45 - cosec60}}}}{{{\text{sec30 + cos60 + cot45}}}}$
As we know sin30 = $\dfrac{1}{2}$, tan45 = $1$, cosec60 = $\dfrac{2}{{\sqrt 3 }}$, sec30 = $\dfrac{2}{{\sqrt 3 }}$, cos60 = $\dfrac{1}{2}$ and cot45 = 1.
$\dfrac{{\dfrac{1}{2} + 1 - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{1}{2} + \dfrac{2}{{\sqrt 3 }} + 1}} = \dfrac{{\dfrac{3}{2} - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{3}{2} + \dfrac{2}{{\sqrt 3 }}}}$
$\dfrac{{\dfrac{3}{2} - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{3}{2} + \dfrac{2}{{\sqrt 3 }}}} = \dfrac{{\dfrac{{3\sqrt 3 - 4}}{{2\sqrt 3 }}}}{{\dfrac{{3\sqrt 3 + 4}}{{2\sqrt 3 }}}} = \dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}$
$\dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}{\text{x}}\dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 - 4}} = \dfrac{{{{(3\sqrt 3 - 4)}^2}}}{{{{(3\sqrt 3 )}^2} - {{(4)}^2}}} = \dfrac{{27 + 16 - 24\sqrt 3 }}{{27 - 16}} = \dfrac{{43 - 24\sqrt 3 }}{{11}}$
(since $(a + b)(a + b) = {(a + b)^2} = {a^2} + {b^2} + 2ab\, and \,(a + b)(a - b) = {a^2} - {b^2}$)
So the correct answer is $\dfrac{{43 - 24\sqrt 3 }}{{11}}$.