# NCERT Solutions Class 10 Ch- 8 Introduction to Trigonometry

### Question 5. Given sec θ = $frac { 13 }{ 12 }$ , calculate all other trigonometric ratios. Solution:

Let us assume a right angled triangle ABC, right angled at B

sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB

According to the Pythagoras theorem,

AC2=AB+ BC2

Substitute the value of AB and AC

(13k)2= (12k)2 + BC2

169k2= 144k2 + BC2

169k2= 144k2 + BC2

BC2 = 169k2 – 144k2

BC2= 25k2

Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios

So,

Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13

Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13

tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12

Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5

cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5

### Question 7.

If cot θ = 7/8, evaluate :

(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)

(ii) cot2 θ

Solution:

Let us assume a △ABC in which ∠B = 90° and ∠C = θ

Given:

cot θ = BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number

According to Pythagoras theorem in △ABC we get.

AC= AB2+BC2

AC= (8k)2+(7k)2

AC= 64k2+49k2

AC= 113k2

AC = √113 k

According to the sine and cos function ratios, it is written as

sin θ = AB/AC = Opposite Side/Hypotenuse = 8k/√113 k = 8/√113 and

cos θ = Adjacent Side/Hypotenuse = BC/AC = 7k/√113 k = 7/√113

Now apply the values of sin function and cos function:

### Question 9.

In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Solution:

### Question 10.

In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P

Solution:

In a given triangle PQR, right angled at Q, the following measures are

PQ = 5 cm

PR + QR = 25 cm

Now let us assume, QR = x

PR = 25-QR

PR = 25- x

According to the Pythagorean Theorem,

PR2 = PQ2 + QR2

Substitute the value of PR as x

(25- x) 2 = 5+ x2

252 + x2 – 50x = 25 + x2

625 + x2-50x -25 – x= 0

-50x = -600

x= -600/-50

x = 12 = QR

Now, find the value of PR

PR = 25- QR

Substitute the value of QR

PR = 25-12

PR = 13

Now, substitute the value to the given problem

(1) sin p = Opposite Side/Hypotenuse = QR/PR = 12/13

(2) Cos p = Adjacent Side/Hypotenuse = PQ/PR = 5/13

(3) tan p =Opposite Side/Adjacent side = QR/PQ = 12/5

11. State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii)cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = 4/3 for some angle θ.

Solution:

(i) The value of tan A is always less than 1.

Proof: In ΔMNC in which ∠N = 90∘,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3 which is greater than.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.

MC2=MN2+NC2

52=32+42

25=9+16

25 = 25

(ii) sec A = 12/5 for some value of angle A

Justification: Let a ΔMNC in which ∠N = 90º,

MC=12k and MB=5k, where k is a positive real number.

By Pythagoras theorem we get,

MC2=MN2+NC2

(12k)2=(5k)2+NC2

NC2+25k2=144k2

NC2=119k2

Such a triangle is possible as it will follow the Pythagoras theorem.

(iii) cos A is the abbreviation used for the cosecant of angle A.

Justification: Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M.

(iv) cot A is the product of cot and A.

Justification: cot M is not the product of cot and M. It is the cotangent of ∠M.

(v) sin θ = 4/3 for some angle θ.

Justification: sin θ = Height/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.

## Ex 8.2

### Question 3.

If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.

Solution:

tan (A + B) = √3

Since √3 = tan 60°

Now substitute the degree value

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equation (i)

tan (A – B) = 1/√3

Since 1/√3 = tan 30°

Now substitute the degree value

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

2A = 90°

A= 45°

Now, substitute the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

## Ex 8.3

### Question 3.

If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:

tan 2A = cot (A- 18°)

We know that tan 2A = cot (90° – 2A)

Substitute the above equation in the given problem

⇒ cot (90° – 2A) = cot (A -18°)

Now, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

Therefore, the value of A = 36°

### Question 4.

If tan A = cot B, prove that A + B = 90°.

Solution:

tan A = cot B

We know that cot B = tan (90° – B)

To prove A + B = 90°, substitute the above equation in the given problem

tan A = tan (90° – B)

A = 90° – B

A + B = 90°

### Question 5.

If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

Solution:

sec 4A = cosec (A – 20°)

We know that sec 4A = cosec (90° – 4A)

To find the value of A, substitute the above equation in the given problem

cosec (90° – 4A) = cosec (A – 20°)

Now, equate the angles

90° – 4A= A- 20°

110° = 5A

A = 110°/ 5 = 22°

Therefore, the value of A = 22°

### Question 6.

If A, B and C are interior angles of a triangle ABC, then show that

sin (B+C/2) = cos A/2

Solution:

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

To find the value of (B+ C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

Now, multiply both sides by sin functions, we get

⇒ sin (B+C)/2 = sin (90°-A/2)

Since sin (90°-A/2) = = cos A/2, the above equation is equal to

sin (B+C)/2 = cos A/2

### Question 7.

Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution:

Given:

sin 67° + cos 75°

In term of sin as cos function and cos as sin function, it can be written as follows

sin 67° = sin (90° – 23°)

cos 75° = cos (90° – 15°)

= sin (90° – 23°) + cos (90° – 15°)

Now, simplify the above equation

= cos 23° + sin 15°

Therefore, sin 67° + cos 75° is also expressed as cos 23° + sin 15°

## Ex 8.4

### Question 4.

Choose the correct option. Justify your choice.
(i) 9 sec2A – 9 tan2A =
(A) 1                 (B) 9              (C) 8                (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0                 (B) 1              (C) 2                (D) – 1
(iii) (sec A + tan A) (1 – sin A) =
(A) sec A           (B) sin A        (C) cosec A      (D) cos A

(iv) 1+tan2A/1+cot2A =

(A) secA                 (B) -1              (C) cot2A                (D) tan2A