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

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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.

Complete step-by-step 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.

On putting the value of the angles in the given equation we get the equation as :

$\dfrac{{\dfrac{1}{2} + 1 - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{1}{2} + \dfrac{2}{{\sqrt 3 }} + 1}} = \dfrac{{\dfrac{3}{2} - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{3}{2} + \dfrac{2}{{\sqrt 3 }}}}$

On solving it further we get,

$\dfrac{{\dfrac{3}{2} - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{3}{2} + \dfrac{2}{{\sqrt 3 }}}} = \dfrac{{\dfrac{{3\sqrt 3 - 4}}{{2\sqrt 3 }}}}{{\dfrac{{3\sqrt 3 + 4}}{{2\sqrt 3 }}}} = \dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}$

We can more simplify it after rationalizing it,

$\dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}{\text{x}}\dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 - 4}} = \dfrac{{{{(3\sqrt 3 - 4)}^2}}}{{{{(3\sqrt 3 )}^2} - {{(4)}^2}}} = \dfrac{{27 + 16 - 24\sqrt 3 }}{{27 - 16}} = \dfrac{{43 - 24\sqrt 3 }}{{11}}$

(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 $\dfrac{{43 - 24\sqrt 3 }}{{11}}$.

Note – Whenever you face such types of problems where the general angles are given directly, put the values of those trigonometric ratios for that particular angle and simplify it further to get the right answer. After solving, we have to do rationalization for removing the square root from the denominator and making the solution more clear. Proceeding like this will take you to the right solution.

Complete step-by-step 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.

On putting the value of the angles in the given equation we get the equation as :

$\dfrac{{\dfrac{1}{2} + 1 - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{1}{2} + \dfrac{2}{{\sqrt 3 }} + 1}} = \dfrac{{\dfrac{3}{2} - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{3}{2} + \dfrac{2}{{\sqrt 3 }}}}$

On solving it further we get,

$\dfrac{{\dfrac{3}{2} - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{3}{2} + \dfrac{2}{{\sqrt 3 }}}} = \dfrac{{\dfrac{{3\sqrt 3 - 4}}{{2\sqrt 3 }}}}{{\dfrac{{3\sqrt 3 + 4}}{{2\sqrt 3 }}}} = \dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}$

We can more simplify it after rationalizing it,

$\dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}{\text{x}}\dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 - 4}} = \dfrac{{{{(3\sqrt 3 - 4)}^2}}}{{{{(3\sqrt 3 )}^2} - {{(4)}^2}}} = \dfrac{{27 + 16 - 24\sqrt 3 }}{{27 - 16}} = \dfrac{{43 - 24\sqrt 3 }}{{11}}$

(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 $\dfrac{{43 - 24\sqrt 3 }}{{11}}$.

Note – Whenever you face such types of problems where the general angles are given directly, put the values of those trigonometric ratios for that particular angle and simplify it further to get the right answer. After solving, we have to do rationalization for removing the square root from the denominator and making the solution more clear. Proceeding like this will take you to the right solution.

Last updated date: 17th Sep 2023

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