Madhya Pradesh Board Question Paper for Class 10th Maths 2017 In PDF

MP Board (MPBSE) Question Paper 2017 Class 10th Mathematics For English Medium with Solutions – Free Download

MP Board Question Paper 2017 Class 10th Maths with solutions can be availed from this page in downloadable pdf format and also in the text, so that learning becomes easier for the students. Maths Question Paper 2017 Class 10 is also added here, which complements the preparation of students within the planned time. Students are able to access all the Madhya Pradesh board previous year maths question papers. The more you practice different types of questions in the previous year papers, the more you succeed in the final board examination.

Question 1: Choose the correct option and write it in your answer book.

(1 * 5 = 5)

Question i: When a₁ / a₂ ≠ b₁ / b₂ then the system of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

(a) Has two solutions (b) Has no solution

(c) Has infinitely many solutions (d) Has unique solution

Answer: (d)

Question ii: In equation 2x + 3y = 6, if y = 0 then write the value of x.

(a) 3 (b) -3 (c) 0 (d) 1 / 3

Answer: (a)

Question iii: The discriminant of the quadratic equation is:

(a) – b² + 4ac (b) b² – 4ac (c) – b² -4ac (d) b² + 4ac

Answer: (b)

Question iv: All circles are:

(a) Congruent (b) Equal (c) Similar (d) All of these

Answer: (c)

Question v: Relation between the arc of the circle, the angle subtended at the centre by arc and radius is:

(a) Arc = radius / angle (b) Radius = angle / arc

(c) Arc = radius + angle (d) Angle = arc / radius

Answer: (d)

Question 2: Write true/false in the following: 1 * 5 = 5

(i) Algebraic expression x² + x√x is a polynomial. [False]

(ii) Education cess is an indirect tax. [True]

(iii) Sum of the opposite angles of a cyclic quadrilateral is 360º. [False]

(iv) All four diagonals of a cuboid are equal. [True]

(v) The arithmetic mean of 7, 8, 9 is 9. [False]

Question 3: Fill in the blanks. 1 * 5 = 5

(i) Zero of a linear polynomial ax + b is ____________ (-b / a)

(ii) Amount paid in installments is always ___________than the cash price or cash down payment. (more)

(iii) Three medians of a triangle are____________ (equal in area)

(iv) The two tangents drawn from an external point to a circle are_____________ (equal)

(v) The probability of impossible event is ___________ (0)

Question 4: Write the answer in on word / sentence of each 1 * 5 = 5

(i) What will be the duplicate ratio of 9:25?

Answer: 9² / 25² = 81 / 625 = 81 . 625

(ii) Write the definition of professional tax.

Answer: Profession tax is the tax levied and collected by the state governments in a country. It is an indirect tax.

(iii) Define the segment of a circle.

Answer: Segment of a circle is a region bounded by a chord of a circle and the intercepted arc of the circle.

(iv) Define a straight line.

Answer: A line is simply an object in geometry that is characterized by a zero-width object that extends on both sides. A straight line is just a line with no curves. So, a line that extends to both sides till infinity and has no curves is called a straight line.

(v) Write the formula of the whole surface of a hollow cylinder.

Answer: Area of a Hollow Cylinder: 2π (r₁ + r₂)( r₁ – r₂ + h)

Question 5: Match the correct column: 1 * 5 = 5

Column A Column B

(i) tan (90^o – θ) (a) √3

(ii) 1 / sec θ (b) ∞

(iii) tan 90^o (c) 1

(iv) tan θ × cosec θ (d) cot θ

(v) sin θ (e) cos θ

(f) √1 – cos² θ

(g) √3 / 2

Answers:

(i) – (d)

(ii) (e)

(iii) – (b)

(iv) – (c)

(v) – (f)

Question 6: [i] Write the characteristic property of side-angle side similarity.

2M

OR

[ii] What are the conditions for the similarity of the triangle?

Solution:

OR

(2) SSS similarity: If the corresponding sides of two triangles are proportional, then the two triangles are similar.

(3) SAS similarity: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

Question 7: [i] In two triangles, the measure of sides and angles are as follows. Show that ∆ABC, ∆PQR are similar or not? AB = 4 centimeter, AC = 5 centimeter, ∠A= 60°, PQ = 6 centimeter, PR = 7.5 centimeter, ∠P = 60°. 2M

OR

[ii] Write the statement of the converse of fundamental proportionality theorem (Thales Theorem).

Solution:

AB / PQ = 4 / 6 = 2 / 3

AC / PR = 5 / 7.5 = 2 / 3

∠A= 60° and ∠P = 60°

Since the two sides and an angle are the same, by SAS similarity of triangles, the triangles ABC and PQR are similar.

OR

Question 8: [i] Check whether 8 cm, 15cm and 17cm are the sides of a right-angled triangle? 2M

OR

[ii] What is the meaning of similar and similarity?

Solution:

Hypotenuse² = Base² + Perpendicular²

17² = 8² + 15²

289 = 64 + 225

289 = 289

LHS = RHS

Thus the given sides form a right triangle.

OR

Question 9: [i] The speeds of 10 Motor Cyclists are recorded in kilometre per hour as follows: 2M

47, 53, 49, 60, 39, 42, 53, 52, 53, 55. Find the mean from the above data.

OR

[ii] Write the names of two methods used to determine the mean of grouped data.

Solution:

= [47 + 53 + 49 + 60 + 39 + 42 + 53 + 52 + 53 + 55] / 10

= 503 / 10

= 50.3

OR

• Direct Mean Method
• Assumed Mean Method
• Step Deviation Mean Method

Question 10: [i] If the probability of raining tomorrow is 0.4 then what will be the probability of not raining tomorrow. 2M

OR

[ii] Find the probability of getting a number greater than 4 in a single throw of dice.

Solution:

P (not raining tomorrow) = 1 – 0.4 = 0.6

OR

P (getting a number greater than 4) = 2 / 6 = 1 / 3

Question 11: [i] Find the length of a chord which is at a distance of 3 centimetres, from the centre of a circle of radius 5 centimetres. 3M

OR

[ii] Prove that the angle in the semicircle is a right angle.

Solution:

In right-angled triangle OMA, by Pythagoras theorem,

OA² = OM² + AM² (The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.)

AM² = 5² – 3² = 16

AM = 4 cm

AB = 2AM

= 2 * 4 cm

= 8 cm

OR

Given: A circle with centre O. PQ is the diameter of the circle subtending ∠PAQ at point A on the circle.

To prove that: ∠PAQ = 90^o

Proof:

POQ is a straight line passing through centre O.

Angle subtended by arc PQ at O is ∠POQ = 180^o —- (1)

Also, by theorem “The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.”

∠POQ = 2∠PAQ

∠POQ / 2 = ∠PAQ

180^o / 2 = ∠PAQ —- (from (1))

90^o = ∠PAQ

∠PAQ = 90^o

Question 12: [i] Find the value of x and y in the given figure: 3M

OR

[ii] Prove that the perpendicular from the centre of a circle to a chord bisects the chord.

Solution:

2x + 10 + 5y – 50 = 180

2x + 5y – 40 = 180

2x + 5y = 220 —– (1)

2x + 4 + 4y – 4 = 180

2x + 4y = 180 —- (2)

Consider the equations (1) and (2)

2x + 5y = 220 —– (1)

2x + 4y = 180 —- (2)

y = 40^o

Substitute y = 40 in (1),

2x + 5 * 40 = 220

2x + 200 = 220

2x = 20

x = 20 / 2

x = 10^o

OR

Given that C is the centre of the circle, AB is a chord such that OX ⟂ AB.

To prove that OX bisects chord AB that is AX = BX.

Proof:

In triangles OAX and OBX,

∠OXA = ∠OXB [Both 90^o]

OA = OB [both radius]

OX = OX [common]

△OAX ⩭ △OBX [by RHS rule]

AX = BX [by CPCT]

Question 13: [i] Compute the mean of the marks obtained in the examination by the students of any class from the following grouped data. 3M

OR

[ii] Find the probability that a leap year, selected randomly, will contain 53 Thursdays.

Solution:

Mean = ∑f_ix_i / ∑f_i

= 2800 / 100

= 28

OR

= > 52 weeks + 2 days.

These 2 days can be:

(a) Monday, Tuesday

(b) Tuesday, Wednesday,

(c) Wednesday, Thursday

(d) Thursday, Friday

(e) Friday, Saturday

(f) Saturday, Sunday

(g) Sunday, Monday.

= > Let A be the event that a leap year contains 53 Thursdays.

n (A) = {Wednesday, Thursday},{Thursday, Friday}}.

= 2

Required probability p (A) = n (A) / n (S)

= 2 / 7

Question 14: [i] Find the mode of the following frequency table: 3M

OR

[ii] Write any 3 uses of the cost of the living index number.

Solution:

Here the maximum frequency is 26, and the class corresponding to this frequency is 40 – 60. So the modal class is 40 – 60.

Therefore, l = 40, h = 10, f₁= 26, f₀ = 17, f₂ = 22

Mode = 40 + {[26 – 17] / [2 * 26 – 17 – 22]} * (10)

= 40 + {(9 / 13)} * 10

= 46.923

OR

(ii) It measures the change in the retail prices of a specified quantity of goods and services.

(iii) It is helpful to the government in framing the policies related to wages.

Question 15: [i] Solve the following system of equations by substitution method. 4M

3x + 2y = 14

-x + 4y = 7

OR

[ii] By adding 5 to the denominator and subtracting 5 from its numerator 1 / 7 is obtained. If 4 is subtracted from its numerator then 1 / 3 is obtained. Find that number.

Solution:

-x + 4y = 7 —- (2)

Multiply equation (2) by 3,

-3x + 12y = 21 —- (3)

3x + 2y = 14 —- (1)

-3x + 12y = 21 —- (3)

14y = 35

y = 35 / 14

y = 5 / 2

Put y = 5 / 2 in (1),

3x + 2 * (5 / 2) = 14

3x + 5 = 14

3x = 9

x = 3

OR

Let the denominator be = y

Therefore, the fraction is = x / y.

Given that,

=> (x – 5) / (y + 5) = 1 / 7

=> (7) (x – 5) = (1) (y + 5)

=> 7x – 35 = y + 5

=> 7x – y = 35 + 5

=> 7x – y = 40 —- (1)

=> (x – 4) / (y) = 1 / 3

=> (3)(x – 4) = y

=> (3x – 12) = y

=> 3x – y = 12 —- (2)

3 * (7x – y = 40)

7 * (3x – y = 12)

21x – 3y = 120 —- (3)

21x – 7y = 84 —-(4)

4y = 36

y = 36 / 4

y = 9

Put y = 9 in (1),

7x – y = 40

7x – 9 = 40

7x = 40 + 9

7x = 49

x = 49 / 7

x = 7

Thus the fraction is 7 / 9.

Question 16: [i] Solve for x and y the system of equations: 4M

x+ ay = b

ax – by = c

OR

[ii] In triangle PQR, ∠P = x°, ∠Q = 3x° and ∠R = y°, 3y – 5x = 30 then find the value of each angle of triangle PQR.

Solution:

ax – by = c ….(ii)

Multiply equation (i) by a,

ax + a²y = ab …. (iii)

Subtracting equation (ii) from equation (iii),

a²y + by = ab – c

y = [ab – c] / [a² + b]

Put the value of y in the equation (i),

x = b – ay

x = b – a {[ab – c] / [a² + b]}

= [b² + ac] / [a² + b]

OR

∠P = x°, ∠Q = 3x° and ∠R = y°

∠P + ∠Q + ∠R = 180^o

x° + 3x° + y° = 180^o

4x° + y° = 180^o

y° = 180^o – 4x° —– (1)

Given,

3y^o – 5x^o = 30^o

3y^o = 30^o + 5x^o —- (2)

Hence, to put the value of 3° in equation (i), multiply equation (i) by 3 on both the sides,

3y^o = 3 (180^o – 4x^o)

30^o + 5x^o = 3 (180^o – 4x^o)

30^o + 5x^o = 540^o – 12x^o

5x^o + 12x^o = 540^o – 30^o

17x^o = 510^o

x^o = 510 / 17 = 30^o

y^o = 180^o – 4x^o

= 180 – 4 * 30

y = 60^o

The three angles of the triangle are:

∠P = x° = 30^o

∠Q = 3x° = 3 * 30^o = 90^o

∠R = y° = 60^o

Question 17: [i] If x = 4ab / [a + b] then prove that [x + 2a] / [x – 2a] + [x + 2b] / [x – 2b] = 2. 4M

OR

[ii] If a / [y + z] + b / [z + x] + c / [x + y] then prove that a (b – c) / y² – z² = b (c – a) / z² – x² = c (a – b) / x² – y².

Solution:

To find [x + 2a] / [x – 2a] + [x + 2b] / [x – 2b]

x = 4ab / [a + b]

=> x = 2a x 2b / a + b

=> x / 2a = 2b / a + b

By componendo dividend,

(x + 2a) / (x – 2a) = (2b + a + b) / (2b – a – b)

Simplify RHS.

(x + 2a) / (x – 2a) = (a + 3b) / (b – a)

Similarly (x + 2b) / (x – 2b) = (3a + b) / (a – b)

(x + 2a) / (x – 2a) + (x + 2b) / (x – 2b)

= (a + 3b) / (b – a) + (3a + b) / (a – b)

= (a + 3b) / (b – a) – (3a + b) / (a – b)

= a + 3b – 3a – b / b – a

= 2b – 2a / b – a

= 2 (b – a) / b – a

= 2

OR

y + z = a / n —- (1)

z + x = b / n —- (2)

x + y = c / n —- (3)

(2) – (1)

x – y = [b – a] / n

(a – b) = -n (x – y)

(3) – (2)

y – z = [c – b] / n

(b – c) = -n (y – z)

(1) – (3)

z – x = [a – c] / n

(c – a) = -n (z – x)

Now consider,

a (b – c) / y² – z²

= a * (-n [y – z] / [y + z] [y – z]

= – na / y + z

= – n * n

= – n²

Similarly,

b (c – a) / z² – x²

= b * (-n [z – x] / [z + x] [z – x]

= – nb / z + x

= – n * n

= – n²

c (a – b) / x² – y²

= c * (-n [x – y] / [y + x] [x – y]

= – nc / x + y

= – n * n

= – n²

Therefore,

a (b – c) / y² – z² = b (c – a) / z² – x² = c (a – b) / x² – y²

Question 18: [i] If ɑ and β are the roots of equations, equation ax² + bx + c = 0, then find the value ɑ² / β + β² / ɑ. 4M

OR

[ii] Solve the following equation by factorization method:

x² – [11x / 4] + [15 / 8] = 0

Solution:

OR

8x² – 22x + 15 / 8 = 0

8x² – 22x + 15 = 0

8x² – 12x − 10x + 15 = 0

4x (2x – 3) – 5 (2x – 3) = 0

(2x – 3) (4x – 5) = 0

2x – 3 = 0 Or 4x – 5 = 0

x = 3 / 2 or x = 5 / 4

Question 19: [i] A building is surmounted by a flag, from a point on the ground 20 meters away from the foot of a building, the angle of elevation of the top of the building and flag are 45° and 60°. Find the height of the building and length of the flag. 4M

OR

[ii] From the top of a 50-meter high hill, the angle of depression of the top and the bottom of a tower is 30° and 45°. Find the height of the tower.

Solution:

tan 45^o = 20 / x

x = 20m = height of the building

tan 60^o = h + 20 / x

√3 = h + 20 / 20

h = 20 (√3 − 1)

Height of the flag = 14.28m

OR

To find the height of the tower CD

Draw BE parallel to AC and DQ.

Lines DQ and BE are parallel, BD is the transversal.

∠DBE = ∠QDB [alternate angles]

∠DBE = 30^o

Lines DQ and AC are parallel, AD is the transversal.

∠DAC = ∠QDA [alternate angles]

∠DAC = 60^o

AC and BE are parallel lines.

AC = BE

Similarly, AB and CE are also parallel lines.

CD = AB

CD = 50m

Since DC is perpendicular to AC, ∠DEB = ∠DCA = 90^o.

tan 30^o = DE / BE

1 / √3 = DE / BE

BE = √3 DE — (1)

tan 60^o = DC / AC

3 = DC / AC

AC = DC / √3 — (2)

From (1) and (2),

√3 DE = DC / √3

DE √3 * √3 = DC

3DE = DC

3DE = DE + EC

3DE – DE = EC

2DE = EC

DE = 50 / 2

DE = 25m

Height of the tower = DC

= DE + EC

= 25 + 50

= 75m

Question 20: [i] Find the length of arc and area of the sectors, if the angles subtended at the centre and the radii are 120° and 2.1cm respectively. 4M

OR

[ii] The radius and height of a cylinder are in the ratio of 2:3. If the volume of the cylinder is 1617 cm³, then find its whole surface area.

Solution:

= [120 / 360] * 2 (22 / 7) * (2.1)

= [0.333] * 13.195

= 4.4cm

Area of the sector = ½ r² θ

= (0.5) * (2.1)² * (120^o)

= 264.6 cm²

OR

Volume of cylinder = πr²h = 1617 cm³

x³ = (1617 * 7) / 22 * 4 * 3 = 343 / 8

Take the cube root of both sides.

x = 7 / 2 = 3.5

So, radius = 2 * 3.5 = 7 cm and the height = 3.5 * 3 = 10.5 cm.

Total Surface area of cylinder = 2πr (r + h)

= 2 * [22 / 7] * 7 (7 + 10.5)

= 2 * 22 * 17.5

= 770 cm²

Question 21: [i] Three solid spheres having the diameters 2 centimetres, 12 centimetres and 16 centimetres respectively are melted to form a sphere. Find the radius of the sphere. 4M

OR

[ii] An iron sphere of diameter 6 centimetres is melted and drawn into a cylindrical wire. If the diameter of the wire is 0.2 centimetre, then find the length of the wire.

Solution:

R³ = 1 + 216 + 512

R³ = 729 = 9³

R = 9cm

The radius of the sphere is 9cm.

OR

For cylindrical wire, d = 0.2cm, radius = 0.2 / 2 = 0.1cm

Volume of cylindrical wire = Volume of the sphere

πr²h = [4 / 3]πr³

π * 0.1² * h = [4 / 3] * π * 3³

h = 113.01 / 0.0314

h = 3600cm or h = 36m

Question 22: [i] Find all other zeros of polynomial f (x) = x³ + 13x² + 32x + 20, if one of the zeros is -2. 5M

OR

[ii] Factorise: a² (b + c) + b² (c + a) + c² (a + b) + 2abc.

Solution:

It is given that -2 is a zero of the function. It means (x + 2) is a factor of a given polynomial.

Divide f (x) by (x + 2), to find the remaining factor.

x³ + 13x² + 32x + 20 / [x + 2] = x² + 11x + 10

The function can be written as

f (x) = (x + 2) (x² + 11x + 10)

f (x) = (x + 2) (x² + 10x + x + 10)

f (x) = (x + 2) (x (x + 10) + 1 (x + 10))

f (x) = (x + 2) (x + 10) (x + 1)

Equate f (x) = 0, to find the zeros.

(x + 2) (x + 10) (x + 1) = 0

x = -2, -10, -1

Therefore, all zeroes of the polynomial are -10, -2 and -1.

OR

= a² b + a²c + b²c + ab²+ c²a + bc² + 2abc

= a (b² + 2bc + c²) + a² b + a²c + b² c + bc²

= a (b + c)² + a² (b + c) + bc (b + c)

= (b + c) [ a² + a (b + c) + bc]

= (b + c) (a + b) (c + a)

Question 23: [i] Solve by using formula 10y² – 11y – 6. 5M

OR

[ii] The sum of the ages of a man and his son is 45 years. Four years ago, the product of their ages was 160. Find their present ages.

Solution:

a = 10, b = -11, c = -6

y = -b ± √b² – 4ac / [2a]

= 11 ± √121 – 4 * (10) * (-6) / [2 * 10]

= 11 ± √121 + 240 / [20]

= 11 ± 19 / 20

y = 11 + 19 / [20]

y = 30 / 20

y = 3 / 2 and

y = 11 – 19 / 2

y = -8 / 2

y = -4

OR

According to question:

x + y = 45 …(1)

Again, 4 years ago, the product of their ages was 160, therefore,

(Age of man 4 years ago) × (Age of son 4 years ago) = 160

(x – 4) × (y – 4) = 160

⇒ (45 – y – 4) × (y – 4) = 160 [Using (1)]

⇒ 45y – 180 – y² + 4y – 4y + 16 = 160

⇒ – y² + 45y – 324 = 0

⇒ y² – 36y – 9y + 324 = 0

⇒ y (y – 36) –9 (y – 36) = 0

⇒ (y – 36) (y – 9) = 0

⇒ y = 36 or y = 9

For y = 36, x = 45 – 36 = 9 which is not possible.

For y = 9, x = 45 – 9 = 36

Question 24: [i] Find the difference between compound interest and simple interest for Rs, 2000 at the rate of 5% per annum interest for 3 years. 5M

OR

[ii] A watch is sold for Rs. 960 cash or for Rs. 480 cash down payment and two monthly instalments of Rs. 245 each. Find the rate of interest charged under the instalment plan.

Solution:

Rate of interest = R = 5%

Time = 3 years

SI = [PTR] / 100

= [2000 * 5 * 3] / 100

= 300

CI = Amount – Principal

A = P [1 + R]^t

A = 2000 [1 + 0.05]³

A = Rs. 2315.25

CI = Amount – Principal

CI = 2315.25 – 2000

CI = 315.25

Difference between SI and CI = 315.25 – 300 = Rs. 15.25.

OR

Cash down payment = Rs. 480

Balance due = 960 − 480 = Rs. 480

Total payment = 245 × 2 = Rs. 490

Total interest paid in instalment

= Rs. 490 − Rs. 480

= Rs. 10

Principal for 1^st month = Rs. 480

Principal for 2^nd month = Rs. 480 − 245

= Rs. 235

Total principal for 1 month = Rs.480 + Rs. 235

= Rs. 715

∴ Rate of interest in the instalment

= [Interest * 100] / [Principal * time]

= [10 × 100] / [715 × (1 / 12)]

= 1000 × 12 / 715

= 16.78%

Question 25: [i] Construct the incircle of the equilateral triangle whose one side is 8 centimetre. Measure its radius. 5M

OR

[ii] Construct a ∆ABC, similar to a given ∆ABC whose sides are in the ratio 7:5.

Solution:

Construction of △ABC –

(1) Draw a line segment BC=8 cm.

(2) Draw an arc of radius =8 cm taking B as the centre.

(3) Draw another arc of radius =8 cm taking B as the centre.

(4) Draw another arc of radius =8 cm. taking C as the centre with intersects the previous arc at a point A.

(5) Join AB and AC.

Thus the require △ABC is constructed.

Construction of the incircle of △ABC.

(1) Draw BP and CQ the bisectors of angles ∠B and ∠C respectively and which intersect each other at point O.

(2) Draw OR⊥BC which intersects BC at L.

(3) Taking O as the centre and OL as radius draw a circle which touches the sides AB, BC and CA at points N, L and M respectively.

(4) Measure the radius OL which is 2.3 cm.

Thus the required incircle is constructed and its radius = 2.3 cm.

OR

Question 26: [i] Simplify the expression (1 + cot θ – cosec θ) (1 + tanθ + sec θ).

5M

OR

[ii] Prove it without using the table: sec 37^o / cosec 53^o + sin 42^o / cos 48^o = 2.

Solution:

(1 + cot θ – cosec θ) (1 + tanθ + sec θ)

Simplifying the given expression,

(1 + cot θ – cosec θ) (1 + tan θ + sec θ)

= (1 + cos θ / sin θ – 1/ sin θ) (1 + sin θ / cos θ + 1 / cos θ)

= (sin θ + cos θ – 1) / sin θ * (cos θ + sin θ + 1) / cos θ

= ((sin θ + cos θ – 1) (cos θ + sin θ + 1)) / (sin θ cos θ)

= ({(sin θ + cos θ) – 1}{(sin θ + cos θ) + 1}) / (sin θ cos θ)

= ((sin θ + cos θ)² – (1)²) / (sin θ cos θ)

= ((sin θ + cos θ)² – (1)²) / (sin θ cos θ)

= ((sin² θ + cos² θ + 2 sin θ cos θ) – 1) / (sin θ cos θ)

= ((sin²θ + cos²θ) + 2 sin θ cos θ – 1) / (sin θ cos θ)

= (1 + 2 sin θ cosθ – 1) / (sinθ cos θ)

= (2 sin θ cos θ) / (sin θ cos θ)

OR

= sec (90 – 53) / cosec 53 + sin (90 – 48) / cos 48

= cosec 53 / cosec 53 + cos 48 / cos 48

= 1 + 1

= 2