Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 13th Jul 2024
Total views: 448.5k
Views today: 8.48k
Verified
448.5k+ views
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 $\dfrac{{43 - 24\sqrt 3 }}{{11}}$.