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

Last updated date: 23rd Mar 2023

•

Total views: 306.9k

•

Views today: 4.84k

Answer

Verified

306.9k+ 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.

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.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE