Question
Evaluate the following:
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
✓
Answer
We have
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ\ \dots(1)$
Now,
$\cot30^\circ=\sqrt{3},\ \cos45^\circ=\frac{1}{\sqrt{2}},\ \sin60^\circ=\frac{\sqrt{3}}{2}$
So by substituting above values in equation (1)
We get,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$=\frac{4}{\big(\sqrt{3}\big)^2}+\frac{1}{\Big(\frac{\sqrt{3}}{2}\Big)^2}-\Big(\frac{1}{\sqrt{2}}\Big)^2$
$=\frac{4}{3}+\frac{1}{\frac{\big(\sqrt{3}\big)^2}{2^2}}-\frac{1^2}{\big(\sqrt{2}\big)^2}$
$=\frac{4}{3}+\frac{2^2}{\big(\sqrt{3}\big)^2}-\frac{1}{2}$
$ =\frac{4}{3}+\frac{4}{3}-\frac{1}{2}$
Now LCM of denominator of above expression is 6
Therefore by taking LCM we get,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$=\frac{4\times2}{3\times2}+\frac{4\times2}{3\times2}-\frac{1\times3}{2\times3}$
$ =\frac{8}{6}+\frac{8}{2}-\frac{3}{6}$
$=\frac{8+8-3}{6}$
$ =\frac{16-3}{6}$
$ =\frac{13}{6}$
Hence,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ=\frac{13}{6}$
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ\ \dots(1)$
Now,
$\cot30^\circ=\sqrt{3},\ \cos45^\circ=\frac{1}{\sqrt{2}},\ \sin60^\circ=\frac{\sqrt{3}}{2}$
So by substituting above values in equation (1)
We get,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$=\frac{4}{\big(\sqrt{3}\big)^2}+\frac{1}{\Big(\frac{\sqrt{3}}{2}\Big)^2}-\Big(\frac{1}{\sqrt{2}}\Big)^2$
$=\frac{4}{3}+\frac{1}{\frac{\big(\sqrt{3}\big)^2}{2^2}}-\frac{1^2}{\big(\sqrt{2}\big)^2}$
$=\frac{4}{3}+\frac{2^2}{\big(\sqrt{3}\big)^2}-\frac{1}{2}$
$ =\frac{4}{3}+\frac{4}{3}-\frac{1}{2}$
Now LCM of denominator of above expression is 6
Therefore by taking LCM we get,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$=\frac{4\times2}{3\times2}+\frac{4\times2}{3\times2}-\frac{1\times3}{2\times3}$
$ =\frac{8}{6}+\frac{8}{2}-\frac{3}{6}$
$=\frac{8+8-3}{6}$
$ =\frac{16-3}{6}$
$ =\frac{13}{6}$
Hence,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ=\frac{13}{6}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solve graphically that the following system of equation has infinitely many solutions:
x - 2y = 5
3x - 6y = 15
Find two consecutive positive even integers whose product is $288$.
→Show that the following points are the vertices of a rectangle:
A(0, -4), B(6, 2), C(3, 5) and D(-3, -1)
A(0, -4), B(6, 2), C(3, 5) and D(-3, -1)
If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)
Two A.P.’s are given 9, 7, 5, . . . and 24, 21, 18, . . . . If nth term of both the progressions are equal then find the value of n and nth term.
Show graphically that the following systems of equations is in-consistent (i.e. has no solution):
$3\text{x}-4\text{y}-1=0$
$2\text{x}-\frac{8}{3}\text{y}+5=0$
→$3\text{x}-4\text{y}-1=0$
$2\text{x}-\frac{8}{3}\text{y}+5=0$
In an annual examination, marks (out of 90) obtained by students of class X in mathematics are given below:
Find the mean marks.
| Marks obtained | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 | 75-90 |
| Number of students | 2 | 4 | 5 | 20 | 9 | 10 |
The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes $\frac{3}{4}.$ Find the fraction.
→The following pie chart gives the marks scored in an examination by a student in various subjects. If the total marks obtained by the student were 360, answer the following questions
(i) Find the marks obtained in each subject.
(ii) How many more marks he got in Mathematics than in Science?
(iii) Which subject he get the least marks?
(i) Find the marks obtained in each subject.
(ii) How many more marks he got in Mathematics than in Science?
(iii) Which subject he get the least marks?
A metallic bucket, open at the top, of height $24\ cm$ is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are $7\ cm$ and $14\ cm,$ respectively. Find
→- The volume of water which can completely fill the bucket,
- The area of the metal sheet used to make the bucket.