JAMB UTME - Mathematics - 1983

Pergunta 1 Relatório

In a sample survey of a University Community, the following table shows the percentage distribution of the number of members per household:

No. of members per household	1	2	3	4	5	6	7	8	Total
No. of households	3	12	15	28	21	10	7	4	100

What is the median?

4

3

5

4.5

none

Detalhes da Resposta

\begin{array}{cc} x & f \\ 1 & 3 \\ 2 & 13 \\ 3 & 15 \\ 4 & 28 \\ 5 & 21 \\ 6 & 10 \\ 7 & 7 \\ 8 & 4 \end{array}

Median is half of total frequency = 50th term
4 falls in the range = 4

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Pergunta 2 Relatório

Factorize completely 81a4 - 16b4

(3a + 2b)(2a - 3b)(9a2 + 4b2)

(3a - 2b)(2a - 3b)(4a2 - 9b2)

(3a - 2b)(3a - 2b)(9a2 - 4b2)

(6a - 2b)(8a - 3b)(4a3 - 9b2)

Detalhes da Resposta

81a4 - 16b4 = (9a2)2 - (4b2)2
= (9a2 + 4b2)(9a2 - 4b2)
N:B 9a2 + 4b2 = (3a - 2b)(3a - 2b)

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Pergunta 3 Relatório

If $ \left(\frac{2}{3}\right)^m \left(\frac{3}{4}\right)^n = \frac{252}{729} $, find the values of m and n

m = 4, n = 2

m = -4, n = -2

m = -4, n =2

m = 4, n = -2

m = -2, n = 4

Detalhes da Resposta

(

\frac{2}{3}

)m (

\frac{3}{4}

)n =

\frac{252}{729}

\frac{2^{m}}{4^{n}}

x

\frac{3^{n}}{3^{m}}

= 283-6
2m - 2n x 3n - m = 28 x 3-6
m - 2n = 8........(i)
-m + n = -6........(ii)
Solving the equations simultaneously
m = 4, n = -2

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Pergunta 4 Relatório

The farm yield of four crops on a piece of land in Ondo are represented on the pie chart. what is the angle of the sector occupies by Okro in the chart?

11

^{\circ}

19

\frac{1}{3}

^{\circ}

33

\frac{1}{3}

^{\circ}

91

\frac{1}{3}

^{\circ}

91

^{\circ}

Detalhes da Resposta

Adding the values of all the items together, it gives 70

Okro sector = $\frac{145}{270}$ x $\frac{360^{o}}{1}$

= 19.33 $^{\circ}$

= 19 $\frac{1}{3}$ $^{\circ}$

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Pergunta 5 Relatório

In a triangle PQT, QR = $ \sqrt{3}\text{ cm} $, PR = 3cm, PQ = $ 2\sqrt{3} $cm and PQR = 30^o. Find angles P and R

P = 60o and R = 90o

P = 30o and R = 120o

P = 90o and R = 60o

P = 60o and R 60o

P = 45o and R = 105o

Detalhes da Resposta

By using cosine formula, p2 = Q2 + R2 - 2QR cos p
Cos P =

\frac{Q^{2} + R^{2} - p^{2}}{2 Q R}

=

\frac{(3)^{2} + 2 (\sqrt{3})^{2} - 3^{2}}{2 \sqrt{3}}

=

\frac{3 + 12 - 9}{12}

=

\frac{6}{12}

=

\frac{1}{2}

= 0.5
Cos P = 0.5
p = cos-1 0.5 = 60

^{\circ}

= < P = 60

^{\circ}

If < P = 60

^{\circ}

and < Q = 30

< R = 180 $^{\circ}$ - 90 $^{\circ}$
angle P = 60 $^{\circ}$ and angle R is 90 $^{\circ}$

Ver Resposta

Pergunta 6 Relatório

The letters of the word MATRICULATION are cut and put into a box. One letter is drawn at random from the box. Find the probability of drawing a vowel

7

13

5

13

6

13

8

13

4

13

Detalhes da Resposta

Vowels of letters are 6 in numbers
prob. of vowel =

\frac{6}{13}

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Pergunta 7 Relatório

Correct each of the numbers 59.81798 and 0.0746829 to three significant figures and multiply them, giving your answer to three significant figures

4.46

4.48

4.47

4.49

4.50

Detalhes da Resposta

59.81798 = 59.8 (3 s.f)
0.0746829 = 0.0747
59.8 x 0.0747 = 4.46706
= 4.48 (3s.f)

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Pergunta 8 Relatório

PQRS is a cyclic quadrilateral. If ∠QPS = 75o, what is the size of ∠QRS?

180o

150o

105o

75o

Circles

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Pergunta 9 Relatório

Simplify $ \frac{3}{2x-1} + \frac{2-x}{x-2} $

2−x

(2x−1)(x−2)

5−x

(2x−1)(x−2)

4−2x

2x−1

6−3x

(2x−1)(x−2)

Detalhes da Resposta

\frac{3}{2 x - 1}

+

\frac{2 - x}{x - 2}

=

\frac{3}{2 x - 1}

-

\frac{x - 2}{x - 2}

=

\frac{3}{2 x - 1}

- 1
=

\frac{3 - (2 x - 1)}{2 x - 1}

=

\frac{3 - 2 x + 1}{2 x - 1}

=

\frac{4 - 2 x}{2 x - 1}

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Pergunta 10 Relatório

Without using tables find the numerical value of $ \log_{7} 49 + \log_{7}\left(\frac{1}{7}\right) $

1

2

3

7

9

Detalhes da Resposta

log749 + log7

\frac{1}{7}

= log77
= 1

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Pergunta 11 Relatório

If M represents the median and D the mode of the measurements 5, 9, 3, 5, 7, 5, 8, then (M, D) is

(6, 5)

(5, 8)

(5, 7)

(5, 5)

(7, 5)

Probability

Linear Inequalities

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Pergunta 12 Relatório

Simplify $ \frac{3\sqrt{27a^{-9}}}{8} $

9a²

2a³

3

2a³

2

3a²

2

Detalhes da Resposta

\frac{3 \sqrt{27 a^{- 9}}}{8}

=

\frac{3 a^{- 3}}{2}

=

\frac{3}{2}

x

\frac{1}{a^{3}}

=

\frac{3}{2 a^{3}}

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Pergunta 13 Relatório

The scores of set of final year students in the first semester examination in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores.

47

50

48

\frac{1}{2}

48

49

Detalhes da Resposta

By re-arranging 21, 23, 29, 40, 41, 43| 47, 50| 50, 55, 56, 70, 70, 73
The median =

\frac{47 + 50}{2}

\frac{97}{2}

= 48

\frac{1}{2}

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Pergunta 14 Relatório

PQRS is a cyclic quadrilateral which PQ = PS. PT is a tangent to the circle and PQ makes an angle of $50^\circ$ with the tangent as shown in the figure. What is the size of QRS?

50

^{\circ}

40

^{\circ}

110v

80

^{\circ}

^{\circ}

Detalhes da Resposta

< SPQ = 80

^{\circ}

< SPQ + < SRQ = 180(Supplementary)

80 + < QRS = 180 $^{\circ}$ - 80 $^{\circ}$

= 100 $^{\circ}$

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Pergunta 15 Relatório

A ship H leaves a port P and sails 30 km due south. Then it sails 50km due west. What is the bearing of H from P

20o 34'

243o 26'

116o 34'

63o 26'

240o 56'

Bearings

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Pergunta 16 Relatório

In the figure PT is a tangent to the circle with centre at O. If PQT = $30^\circ$, find the value of PTO

30

^{\circ}

50

^{\circ}

24

^{\circ}

12

^{\circ}

60

^{\circ}

Detalhes da Resposta

FROM the diagram, PQT = 50

^{\circ}

PTQ = 50 $^{\circ}$ (opposite angles are supplementary)

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Pergunta 17 Relatório

A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5. At the end of the year, Y received ₦5,000 more than X. What is the total profit of the company for the year?

₦20, 000

₦25,000

₦30,000

₦15,000

₦45, 000

Detalhes da Resposta

Total sharing ratio is 9
X has 4, Y has 4 + 1
If 1 is ₦5000
Total profit = 5000 x 9
= ₦45,000

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Pergunta 18 Relatório

In the figure, find PRQ

66

\frac{1}{2}

^{\circ}

^{\circ}

125

^{\circ}

105

^{\circ}

65

^{\circ}

Detalhes da Resposta

Angle subtended at any part of the circumference of the circle

\frac{125^{o}}{2}

at centre = 360

^{\circ}

- 235

^{\circ}

= 125

^{\circ}

$\bar{P Q R}$ = $\frac{125}{2}$

= 62 $\frac{1}{2}$ $^{\circ}$

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Pergunta 19 Relatório

Simplify $ \log_{10} a^{\frac{1}{3}} + \frac{1}{4}\log_{10} a - \frac{1}{12}\log_{10} a^{7} $

1

\frac{7}{6}

log10 a

zero

10

Detalhes da Resposta

log10 a

\frac{1}{3}

+

\frac{1}{4}

log10 a -

\frac{1}{12}

log10a7 = log10 a

\frac{1}{3}

+ log10

\frac{1}{4}

- log10 a

\frac{7}{12}

= log10 a

\frac{7}{12}

- log10 a

\frac{7}{12}

= log10 1 = 0

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Pergunta 20 Relatório

In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm. Determine PS

2

\sqrt{3}

cm

4

\sqrt{6}

cm

2

\sqrt{6}

cm

8

\sqrt{6}

cm

8cm

Detalhes da Resposta

From the diagram, sin 60o =

\frac{P R}{8}

PR = 8 sin 60 =

\frac{8 \sqrt{3}}{2}

= 4

\sqrt{3}

Cos 45o =

\frac{P R}{P S}

=

\frac{4 \sqrt{3}}{P S}

PS Cos45o = 4

\sqrt{3}

PS = 4

\sqrt{3}

x 2
= 4

\sqrt{6}

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Pergunta 21 Relatório

If $0.0000152 \times 0.042 = A \times 10^8$, where $1 \le A < 10$, find A and B

A = 9, B = 6.38

A = 6.38, B = -9

A = -6.38, B = -9

A = -9, B = -6.38

Detalhes da Resposta

0.0000152 x 0.042 = A x 108

1 $\leq$ A < 10, it means values of A includes 1 - 9

0.0000152 = 1.52 x 10-5

0.00042 = 4.2 x 10-4

1.52 x 4.2 = 6.384

10-5 x 10-4

= 10-5-4

= 10-9

= 6.38 x 10-9

A = 6.38, B = -9

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Pergunta 22 Relatório

Make T the subject of the equation $ \frac{av}{1-v} = \sqrt{\frac{2v+T}{a+2T}} $

T = $\frac{3 a v}{1 - v}$

T = $\frac{1 + v}{2 a^{2} v^{3}}$

T = $\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} + (1 - v)^{2}}$

\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} - (1 - v)^{3}}

Detalhes da Resposta

$\frac{a v}{1 - v}$ = $\sqrt{\frac{2 v + T}{a + 2 T}}$
$\frac{(a v)^{3}}{(1 - v)^{3}}$ = $\frac{2 v + T}{a + 2 T}$
$\frac{a^{3} v^{3}}{(1^{3} - v)^{3}}$ = $\frac{2 v + T}{a + 2 T}$
= $\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} - (1 - v)^{3}}$

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Pergunta 23 Relatório

Find x if (xbase4)2 = 100100base2

6

12

100

210

110

Detalhes da Resposta

(x2)4 to base10 gives (1 x 25) + (1 x 25) + (1 x 2)
32 + 4 = 36
x2 = 36, x = 6
610 to base 4 =

\begin{array}{cc} 4 & 6 \\ 1 r 2 \end{array}

= 124

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Pergunta 24 Relatório

If O is the centre of the circle in the figure, find the value of x

50

260

100

65

130

Detalhes da Resposta

From the diagram; The value of x = 360

^{\circ}

- 2(130

^{\circ}

)

= 360 - 260

= 100 $^{\circ}$

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Pergunta 25 Relatório

Solve the following equations 4x - 3 = 3x + y = x - y = 3, 3x + y = 2y + 5x - 12

x = 5, y = 2

x = 2, y = 5

x = 5, y = -2

x = -2, y = -5

x = -5, y = -2

Detalhes da Resposta

4x - 3 = 3x + y = x - y = 3.......(i)
3x + y = 2y + 5x - 12.........(ii)
eqn(ii) + eqn(i)
3x = 15
x = 5
substitute for x in equation (i)
5 - y = 3
y = 2

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Pergunta 26 Relatório

Given a regular hexagon, calculate each interior angle of the hexagon

60o

30o

120o

45o

135o

Detalhes da Resposta

Sum of interior angles of polygon = (2n - 4)rt < s

sum of interior angles of an hexagon
(2 x 6 - 4) x 90o = (12 - 4) x 90o
= 8 x 90o
= 720o
each interior angle will have $\frac{720^{o}}{6}$
= 120o

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Pergunta 27 Relatório

PQRS is a desk of dimensions $2\text{m} \times 0.8$ which is inclined at $30^\circ$ to the horizontal. Find the inclination of the diagonal PR to the horizontal

30

^{\circ}

35'

30

^{\circ}

32'

15

^{\circ}

36'

10

^{\circ}

42'

Detalhes da Resposta

tan

θ

=

\frac{0.8}{2}

= (0.4)

$θ$ = tan-1(0.4)

From the diagram, the inclination of the diagonal PR to the horizontal is 10 $^{\circ}$ 42'

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Pergunta 28 Relatório

In a class of 60 pupils, the statistical distribution of the numbers of pupils offering Biology, History, French, Geography and Additional mathematics is as shown in the pie chart. How many pupils offer Additional Mathematics?

15

10

18

12

20

Detalhes da Resposta

2x - 24)

^{\circ}

+ (3x - 18)

^{\circ}

+ (2x + 12)

^{\circ}

+ (x + 12)

^{\circ}

+ x

^{\circ}

= 360

^{\circ}

9x = 360 $^{\circ}$ + 18 $^{\circ}$

x = $\frac{378}{9}$

= 42 $^{\circ}$ , if x = 42 $^{\circ}$ , then add maths = $\frac{2 x - 42}{360}$ x 60

= $\frac{2 \times 42 - 24}{360}$ x 60

= $\frac{84 - 24}{6}$

= 10

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Pergunta 29 Relatório

Find the angle of the sectors representing each item in pie chart of the following data 6, 10, 14, 16, 26

15o, 25o, 35o, 40o, 65o

60o, 100o, 140o, 160o, 260o

6o, 10o, 14o, 16o, 26o

30o, 50o, 70o, 80o, 130o

none of the above

Detalhes da Resposta

6 + 10 + 14 + 16 + 26 = 72

\frac{6}{72}

x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively

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Pergunta 30 Relatório

Solve the simultaneous equations for x in x2 + y - 8 = 0, y + 5x - 2 = 0

-28, 7

6, -28

6, -1

-1, 7

3, 2

Detalhes da Resposta

x2 + y - 8 = 0, y + 5x - 2 = 0
Rearranging, x2 + y = 8.....(i)
5x + y = 2.......(ii)
Subtract eqn(ii) from eqn(i)
x2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6, -1

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Pergunta 31 Relatório

If x + 2 and x - 1 are factors of the expression 1x3 + 2kx2 + 24, find the values of 1 and K

1 = -6, k = -9

1 = -2, k = 1

k = -1, 1 = -2

1 = 0, k = 1

k = 0,1 = 6

Detalhes da Resposta

f(x) = Lx3 + 2kx2 + 24
f(-2) = -8L + 8k = -24
4L - 4k = 12
f(1):L + 2k = -24
L - 4k = 3
3k = -27
k = -9
1 = -6

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Pergunta 32 Relatório

The scores of 16 students in a mathematics test are 65, 65, 55, 60, 60, 65, 60, 70, 75, 70, 65, 70, 60, 65, 65,
70. What is the sum of the median and modal scores?

125

130

140

150

Measures Of Dispersion

Statistics

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Pergunta 33 Relatório

Find the mean of the following 24.57, 25.63, 24.32, 26.01, 25.77

25.12

25.30

25.26

25.50

25.75

Detalhes da Resposta

\frac{24.57 + 25.63 + 24.32 + 26.01 + 25.77}{5}

mean =

\frac{126.3}{5}

= 25.26

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Pergunta 34 Relatório

The figure FGHK is a rhombus. What is the value of angle X?

30

^{\circ}

90

^{\circ}

150

^{\circ}

120

^{\circ}

60

^{\circ}

Detalhes da Resposta

< HKF = 60

^{\circ}

, < KFG = 120

^{\circ}

< KFG = < KHG = x(opposite angles)

x = 120 $^{\circ}$

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Pergunta 35 Relatório

A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Although he covered 600 kilometers. At what speed did he drive for the last 3 hours?

120km/hr

60km/hr

670km/hr

40km/hr

Detalhes da Resposta

Speed =

\frac{d i s t a n c e}{t i m e}

let x represent the speed, d represent distance
x =

\frac{d}{4}

d = 4x
2x =

\frac{600 - d}{3}

6x = 600 - d
6x = 600 - 4x
10x = 600
x =

\frac{600}{10}

= 60km/hr

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Pergunta 36 Relatório

PQR is the diagram of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = 60^o. Find the perimeter of the figure PTRS. $ \pi = \frac{22}{7} $

25cm

18cm

36cm

29cm

25.5cm

Detalhes da Resposta

Circumference of PRS =

\frac{π}{2}

=

\frac{22}{7}

x

\frac{7}{1}

x

\frac{1}{2}

= 11cm
Side PT = 7cm, Side TR = 7cm
Perimeter(PTRS) = 11cm + 7cm + 7cm
= 25cm

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Pergunta 37 Relatório

If a rod of length 250cm is measured as 255cm long in error, what is the percentage error in the measurement?

55

10

5

4

2

Detalhes da Resposta

% error =

\frac{Actual error}{real value}

x 100
=

\frac{5}{250}

x 100
= 2

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Pergunta 38 Relatório

One interior angle of a convex hexagon is 170o and each of the remaining interior angles is equal to xo. Find x

120o

110o

105o

102o

100o

Detalhes da Resposta

An hexagon polygon is a six sided polygon
n = 6
sum of all interior angles of hexagon polygon will be (2n - 4) x 90o
= [2 x 6 - 4] x 90
= 720o
If one angle is 170o and each of the remaining five angles is 5xo
5xo + 170o = 720o
= 5x + 550
x =

\frac{550}{5}

= 110o

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Pergunta 39 Relatório

If x is jointly proportional to the cube of y and the fourth power of z. In what ratio is x increased or decreased when y is halved and z is doubled?

4:1 increase

2:1 increase

1:1 no change

1:4 decrease

3:4 decrease

Variation

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Pergunta 40 Relatório

Find the missing value in the following table:

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y=x^3-x+3$		$3$	$3$	$3$	$9$	$27$

-3

3

-9

13

9

Detalhes da Resposta

When x = -2, y = x3 - x + 3
= -23 - (-2) + 3
= -8 + 2 + 3
= -3

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Pergunta 41 Relatório

The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x

2

6

18

12

5

Detalhes da Resposta

(3x + 1)2 = (3x - 1)2 + 2 (Pythagoras's theorem)
9x2 + 6x + 1 = 9x - 6x2 + 1 + x2
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12

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Pergunta 42 Relatório

If w varies inversely as V and U varies directly as w3, Find the relationship between u and v given that u = 1, when v = 2

u =

\frac{8}{v^{3}}

v =

\frac{8}{u^{2} v^{3}}

u = 8v3

v = 8u2

Detalhes da Resposta

W

α

\frac{1}{v}

u

α

w3
w =

\frac{k_{1}}{v}

u = k2w3
u = k2(

\frac{k_{1}}{v}

)3
=

\frac{k_{2} k_{1}^{2}}{v^{3}}

k = k2k1k2
u =

\frac{k}{v^{3}}

k = uv3
= (1)(2)3
= 8
u =

\frac{8}{v^{3}}

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Variation

Change Of Subject Of A Formula/Relation

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Pergunta 43 Relatório

The value of $ (0.303)^3 - (0.02)^3 $ is

0.019

0.0019

0.00019

0.000019

0.0278

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Pergunta 44 Relatório

Which of the following equations represents the graph above?

y = 2 + 7x + 4x2

y = 2 - 7x + 4x2

y = 2 + 7x - 4x2

y = 2 - 7x - 4x2

Detalhes da Resposta

x = -2 or x =

\frac{1}{4}

(x + 2)(4x - 1) = 0

y = 2 - 7x - 4x2

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Pergunta 45 Relatório

y varies partly as the square of x and partly as the inverse of the square root of x. Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4.

y = $\frac{10 x^{2}}{31} + \frac{52}{31 \sqrt{x}}$

y = x2 + $\frac{1}{\sqrt{x}}$

y = x2 + $\frac{1}{x}$

y = $\frac{x^{2}}{31} + \frac{1}{31 \sqrt{x}}$

Detalhes da Resposta

y = kx2 + $\frac{c}{\sqrt{x}}$
y = 2 when x = 1
2 = k + $\frac{c}{1}$
k + c = 2
y = 6 when x = 4
6 = 16k + $\frac{c}{2}$
12 = 32k + c
k + c = 2
32k + c = 12
= 31k + 10
k = $\frac{10}{31}$
c = 2 - $\frac{10}{31}$
= $\frac{62 - 10}{31}$
= $\frac{52}{31}$
y = $\frac{10 x^{2}}{31} + \frac{52}{31 \sqrt{x}}$

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Pergunta 46 Relatório

In the figure, PQR is a straight line. Find the values of x and y.

^{\circ}

x = 15

^{\circ}

, y = 52.5

^{\circ}

x = 22.5

^{\circ}

, y = 45.0

^{\circ}

x = 56.25

^{\circ}

, y = 11.5

^{\circ}

Detalhes da Resposta

\frac{2}{3}

+ 3y + 45

^{\circ}

= 180

^{\circ}

3x + 6y + 90 $^{\circ}$ = 360 $^{\circ}$

3x + 67 = 270.......(i) x 2, 5x + y + y = 180 $^{\circ}$

5x + 2y = 180 $^{\circ}$ .......(ii) x 6

6 x 12y = 540..........(iii)

30x + 12y = 1080.........(iv)

egn(iv) - eqn(iii)

24x = 540

x = 22.5 and y = 33.75 $^{\circ}$

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Pergunta 47 Relatório

On a square paper of length 2.524375cm is inscribed square diagram of length 0.524375cm. Find the area of the paper not covered by the diagram. correct to 3 significant figures.

6.00cm2

6.10cm2

6cmv

6.09cm2

4.00cm2

Detalhes da Resposta

Area of the paper = area of square = L x B or S2
where s = S x k
Area of the paper = (2.524)2
area of the diagram = (0.524)2
area not covered = (2.524)2 - (0.524)2
= 6.370576 - 0.274576
= 6.096
= 6.10cm2 (2 s.f)

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Coordinate Geometry

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Pergunta 48 Relatório

Given that cos z = L , whrere z is an acute angle, find an expression for $ \frac{\cot z - \cosec z}{\sec z + \tan z} $

1-L

1+L

L²√3

1+L

1+L³

L²

L(L−1)

1−L+1√1−L²

Detalhes da Resposta

Given Cos z = L, z is an acute angle
$\frac{cot z - cosec z}{sec z + tan z}$ = cos z
= $\frac{cos z}{sin z}$
cosec z = $\frac{1}{sin z}$
cot z - cosec z = $\frac{cos z}{sin z}$ - $\frac{1}{sin z}$
cot z - cosec z = $\frac{L - 1}{sin z}$
sec z = $\frac{1}{cos z}$
tan z = $\frac{sin z}{cos z}$
sec z = $\frac{1}{cos z}$ + $\frac{sin z}{cos z}$
= $\frac{1}{l}$ + $\frac{sin z}{L}$
the original eqn. becomes
$\frac{cot z - cosec z}{sec z + tan z}$ = $\frac{L - \frac{1}{sin z}}{1 + s i n \frac{z}{L}}$
= $\frac{L (L - 1)}{sin z (1 + sin z)}$
= $\frac{L (L - 1)}{sin z + 1 - c o s^{2} z}$
= sin z + 1
= 1 + $\sqrt{1 - L^{2}}$
= $\frac{L (L - 1)}{1 - L + 1 \sqrt{1 - L^{2}}}$

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Matrices And Determinants

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\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)
\(y=x^3-x+3\)		\(3\)	\(3\)	\(3\)	\(9\)	\(27\)






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