# NCERT Solutions Class 10 for Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions Class 10 for Maths Chapter 8 Introduction to Trigonometry. : In this post, we will share with you all the detailed NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry.. This will contain both in-text and back-exercise questions for Science and Social Science, and all exercise questions for Mathematics. For all school and board level examinations, doing all the NCERT Questions is a must.

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# Class 10 Maths Chapter 8 Introduction to Trigonometry

## Exercise 8.1 (Page 181)

Question 1:
In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A
(ii) sin C, cos C

In a given triangle ABC, right angled at B = ∠B = 90°
Given: AB = 24 cm and BC = 7 cm
According to the Pythagoras Theorem,
In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying Pythagoras theorem, we get
AC2=AB2+BC2
AC2 = (24)2+72
AC2 = (576+49)
AC2 = 625cm2
AC = √625 = 25

Therefore, AC = 25 cm

(i) To find Sin (A), Cos (A)
We know that sine (or) Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes

Sin (A) = Opposite side /Hypotenuse = BC/AC = 7/25
Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,
Cos (A) = Adjacent side/Hypotenuse = AB/AC = 24/25

(ii) To find Sin (C), Cos (C)
Sin (C) = AB/AC = 24/25
Cos (C) = BC/AC = 7/25

Question 2:
In Fig. 8.13, find tan P – cot R

In the given triangle PQR, the given triangle is right angled at Q and the given measures are:

PR = 13cm,
PQ = 12cm

Since the given triangle is right angled triangle, to find the side QR, apply the Pythagorean theorem
According to Pythagorean theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.
PR2 = QR2 + PQ2

Substitute the values of PR and PQ

13= QR2+122
169 = QR2+144

Therefore, QR= 169−144
QR= 25
QR = √25 = 5

Therefore, the side QR = 5 cm
To find tan P – cot R:

According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes

tan (P) Opposite side /Adjacent side = QR/PQ = 5/12
Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,
Cot (R) = Adjacent side/Opposite side = QR/PQ = 5/12

Therefore,

tan (P) – cot (R) = 5/12 – 5/12 = 0
Therefore, tan(P) – cot(R) = 0

Question 3:
If sin A = 3/4, Calculate cos A and tan A.

Let us assume a right angled triangle ABC, right angled at B
Given: Sin A = 3/4

We know that, Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.
Therefore, Sin A = Opposite side /Hypotenuse= 3/4

Let BC be 3k and AC will be 4k
where k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC2=AB2 + BC2

Substitute the value of AC and BC
(4k)2=AB2 + (3k)2
16k2−9k2 =AB2
AB2=7k2

Therefore, AB = √7k
Now, we have to find the value of cos A and tan A
We know that,

Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get
AB/AC = √7k/4k = √7/4

Therefore, cos (A) = √7/4
Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

BC/AB = 3k/√7k = 3/√7
Therefore, tan A = 3/√7

Question 4:
Given 15 cot A = 8, find sin A and sec A.

Let ΔABC be a right-angled triangle, right-angled at B.We know that cot A = AB/BC = 8/15   (Given)
Let AB be 8k and BC will be 15k where k is a positive real number. By Pythagoras theorem we get,
AC2 = AB2 + BC
AC2 = (8k)2 + (15k)2
AC2 = 64k2 + 225k2
AC2=289k2
AC = 17 k
sin A = BC/AC = 15k/17k = 15/17sec A = AC/AB = 17k/8 k = 17/8

Question 5:
Given sec θ = 13/12 Calculate all other trigonometric ratios.

We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side

Let us assume a right angled triangle ABC, right angled at B
sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB

Let AC be 13k and AB will be 12k
Where, k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,
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 6:
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Let ΔABC in which CD ⊥ AB.A/q,
cos A = cos B⇒ AD/AC = BD/BC⇒ AD/BD = AC/BCLet AD/BD = AC/BC = k⇒ AD = kBD  …. (i)⇒ AC = kBC  …. (ii)By applying Pythagoras theorem in ΔCAD and ΔCBD we get,
CD2 = AC2 – AD2 …. (iii)
and also CD2 = BC2 – BD2 …. (iv)
From equations (iii) and (iv) we get,
AC2 – AD2 = BC2 – BD2
⇒ (kBC)2 – (k BD)2 = BC2 – BD2
⇒ k2 (BC2 – BD2) = BC2 – BD2
⇒ k2 = 1
⇒ k = 1
Putting this value in equation (ii), we obtain
AC = BC
⇒ ∠A = ∠B  (Angles opposite to equal sides of a triangle are equal-isosceles triangle)

Question 7:
If cot θ = 7/8, evaluate :

(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)
(ii) cot2 θ

Let ΔABC in which ∠B = 90º and ∠C = θA/q,cot θ = BC/AB = 7/8
Let BC = 7k and AB = 8k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC
AC2 = (8k)2 + (7k)2
AC2 = 64k2 + 49k2
AC2=113k2
AC = √113 k
sin θ = AB/AC = 8k/√113 k = 8/√113 and cos θ = BC/AC = 7k/√113 k = 7/√113
(i) (1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ) = (1-sin2θ)/(1-cos2θ) = {1 – (8/√113)2}/{1 – (7/√113)2}     = {1 – (64/113)}/{1 – (49/113)} = {(113 – 64)/113}/{(113 – 49)/113} = 49/64

(ii) cot2θ = (7/8)2 = 49/64

Question 8:
If 3 cot A = 4, check whether (1-tan2 A)/(1+tan2 A) = cos2 A – sin 2 A or not

Let ΔABC in which ∠B = 90º,
A/q,
cot A = AB/BC = 4/3
Let AB = 4k and BC = 3k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC
AC2 = (4k)2 + (3k)2
AC2 = 16k2 + 9k2
AC2=25k2
AC = 5k
tan A = BC/AB = 3/4
sin A = BC/AC = 3/5
cos A = AB/AC = 4/5
L.H.S. = (1-tan2A)/(1+tan2A) = 1- (3/4)2/1+ (3/4)= (1- 9/16)/(1+ 9/16) = (16-9)/(16+9) = 7/25
R.H.S. = cos2A – sin2A = (4/5)– (3/4)2 =(16/25) – (9/25) = 7/25
R.H.S. = L.H.S.
Hence,  (1-tan2A)/(1+tan2A) = cos2A – sin2A

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

Let ΔABC in which ∠B=90°
tan A = BC/AB = 1/√3
Let BC = 1k and AB = √3 k,

Where k is the positive real number of the problem
By Pythagoras theorem in ΔABC we get:

AC2=AB2+BC2
AC2=(√3 k)2+(k)2
AC2=3k2+k2
AC2=4k2
AC = 2k

Now find the values of cos A, Sin A

Sin A = BC/AC = 1/2
Cos A = AB/AC = √3/2

Then find the values of cos C and sin C

Sin C = AB/AC = 3/2
Cos C = BC/AC = 1/2

Now, substitute the values in the given problem

(i) sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1

(ii) cos A cos C – sin A sin C = (3/2 )(1/2) – (1/2) (3/2 ) = 0

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

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

Question 11:

(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.

(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.

## Exercise 8.2 (Page 187)

Question 1:
1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan245° + cos230° – sin260°
(iii) cos 45°/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)

(i) sin 60° cos 30° + sin 30° cos 60°
=  (√3/2×√3/2) + (1/2×1/2) = 3/4 + 1/4 = 4/4 = 1

(ii) 2 tan245° + cos230° – sin260°
= 2×(1)+ (√3/2)2 – (√3/2)= 2

(iii) cos 45°/(sec 30° + cosec 30°)
= 1/√2/(2/√3 + 2) = 1/√2/{(2+2√3)/√3)
= √3/√2×(2+2√3) = √3/(2√2+2√6)
= √3(2√6-2√2)/(2√6+2√2)(2√6-2√2)
= 2√3(√6-√2)/(2√6)2-(2√2)2
=  2√3(√6-√2)/(24-8) =  2√3(√6-√2)/16
= √3(√6-√2)/8 = (√18-√6)/8 = (3√2-√6)/8

(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
= (1/2+1-2/√3)/(2/√3+1/2+1)
= (3/2-2/√3)/(3/2+2/√3)
= (3√3-4/2√3)/(3√3+4/2√3)
= (3√3-4)/(3√3+4)
= (3√3-4)(3√3-4)/(3√3+4)(3√3-4)
= (3√3-4)2/(3√3)2-(4)2
= (27+16-24√3)/(27-16)
= (43-24√3)/11]

(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)
= 5(1/2)2+4(2/√3)2-12/(1/2)2+(√3/2)2
= (5/4+16/3-1)/(1/4+3/4)
= (15+64-12)/12/(4/4)
= 67/12

Question 2:
Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan230° =
(A) sin 60°            (B) cos 60°          (C) tan 60°            (D) sin 30°
(ii) 1-tan245°/1+tan245° =
(A) tan 90°            (B) 1                    (C) sin 45°            (D) 0
(iii)  sin 2A = 2 sin A is true when A =
(A) 0°                   (B) 30°                  (C) 45°                 (D) 60°
(iv) 2tan30°/1-tan230° =
(A) cos 60°          (B) sin 60°             (C) tan 60°           (D) sin 30°

(i) (A) is correct.
2tan 30°/1+tan230° = 2(1/√3)/1+(1/√3)2
= (2/√3)/(1+1/3) = (2/√3)/(4/3)= 6/4√3 = √3/2 = sin 60°

(ii)  (D) is correct.1-tan245°/1+tan245° = (1-12)/(1+12)= 0/2 = 0

(iii) (A) is correct.sin 2A = 2 sin A is true when A == As sin 2A = sin 0° = 0
2 sin A = 2sin 0° = 2×0 = 0or,sin 2A = 2sin A cos A⇒2sin A cos A = 2 sin A
⇒ 2cos A = 2 ⇒ cos A = 1⇒ A = 0°

(iv) (C) is correct.2tan30°/1-tan230° =  2(1/√3)/1-(1/√3)2= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°

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

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°

Question 4:

(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.

(i) False.
Let A = 30° and B = 60°, then
sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,
sin A + sin B = sin 30° + sin 60°= 1/2 + √3/2 = 1+√3/2

(ii) True.sin 0° = 0sin 30° = 1/2
sin 45° = 1/√2sin 60° = √3/2
sin  90° = 1
Thus the value of sin θ increases as θ increases.

(iii) False.cos 0° = 1cos 30° = √3/2
cos 45° = 1/√2
cos 60° = 1/2
cos 90° = 0Thus the value of cos θ decreases as θ increases.

(iv) True.cot A = cos A/sin A
cot 0° = cos 0°/sin 0° = 1/0 = undefined.

## Exercise 8.3 (Page 189)

Question 1:
Evaluate :

(i) sin 18°/cos 72°
(ii) tan 26°/cot 64°
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°

(i) sin 18°/cos 72°
To simplify this, convert the sin function into cos function
We know that, 18° is written as 90° – 18°, which is equal to the cos 72°.
= sin (90° – 18°) /cos 72°

Substitute the value, to simplify this equation
= cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°
To simplify this, convert the tan function into cot function
We know that, 26° is written as 90° – 36°, which is equal to the cot 64°.

= tan (90° – 36°)/cot 64°

Substitute the value, to simplify this equation
= cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°
To simplify this, convert the cos function into sin function
We know that, 48° is written as 90° – 42°, which is equal to the sin 42°.

= cos (90° – 42°) – sin 42°

Substitute the value, to simplify this equation
= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°
To simplify this, convert the cosec function into sec function
We know that, 31° is written as 90° – 59°, which is equal to the sec 59°

= cosec (90° – 59°) – sec 59°

Substitute the value, to simplify this equation
= sec 59° – sec 59° = 0

Question 2:
Show that:

(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0

(i) tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°
= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°
= sin 52° sin 38° – sin 38° sin 52° = 0

Question 3:
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

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°.

A/q, tan A = cot B
⇒ 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.

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

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

Hence proved.

Question 7:
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

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°

## Exercise 8.4 (Page 193)

Question 1:
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas
We know that,

cosec2A– cot2A = 1
cosec2A = 1 + cot2A

Since cosec function is the inverse of sin function, it is written as
1/sin2A = 1 + cot2A
Now, rearrange the terms, it becomes
sin2A = 1/(1+cot2A)

Now, take square roots on both sides, we get
sin A = ±1/(√(1+cot2A)

The above equation defines the sin function in terms of cot function
Now, to express sec function in terms of cot function, use this formula
sin2A = 1/ (1+cot2A)

Now, represent the sin function as cos function
1 – cos2A = 1/ (1+cot2A)

Rearrange the terms,
cos2A = 1 – 1/(1+cot2A)
⇒cos2A = (1-1+cot2A)/(1+cot2A)

Since sec function is the inverse of cos function,
⇒ 1/sec2A = cot2A/(1+cot2A)

Take the reciprocal and square roots on both sides, we get
⇒ sec A = ±√ (1+cot2A)/cotA

Now, to express tan function in terms of cot function
tan A = sin A/cos A and cot A = cos A/sin A

Since cot function is the inverse of tan function, it is rewritten as
tan A = 1/cot A

Question 2:
Write all the other trigonometric ratios of ∠A in terms of sec A.

Cos A function in terms of sec A:
sec A = 1/cos A

⇒ cos A = 1/sec A

sec A function in terms of sec A:
cos2A + sin2A = 1

Rearrange the terms

sin2A = 1 – cos2A
sin2A = 1 – (1/sec2A)
sin2A = (sec2A-1)/sec2A
sin A = ± √(sec2A-1)/sec A

cosec A function in terms of sec A:
sin A = 1/cosec A
⇒cosec A = 1/sin A
cosec A = ± sec A/√(sec2A-1)

Now, tan A function in terms of sec A:
sec2A – tan2A = 1

Rearrange the terms
⇒ tan2A = sec2A – 1
tan A = √(sec2A – 1)
cot A function in terms of sec A:
tan A = 1/cot A
⇒ cot A = 1/tan A
cot A = ±1/√(sec2A – 1)

Question 3:
Evaluate:

(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii) sin 25° cos 65° + cos 25° sin 65°

(i) (sin263° + sin227°)/(cos217° + cos273°)
= [sin2(90°-27°) + sin227°]/[cos2(90°-73°) + cos273°)]
= (cos227°+ sin227°)/(sin227° + cos273°)
= 1/1 =1                       (∵ sin2A + cos2A = 1)

(ii) sin 25° cos 65° + cos 25° sin 65°
= sin(90°-25°) cos 65° + cos(90°-65°) sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= cos265°+ sin265° = 1

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) (secA + tanA) (1 – sinA) =
(A) secA           (B) sinA        (C) cosecA      (D) cosA
(iv) 1+tan2A/1+cot2A =
(A) sec2A                 (B) -1              (C) cot2A                (D) tan2A

(i) (B) is correct.
Justification:
Take 9 outside, and it becomes

9 sec2A – 9 tan2A
= 9 (sec2A – tan2A)
= 9×1 = 9             (∵ sec2 A – tan2 A = 1)

Therefore, 9 sec2A – 9 tan2A = 9

(ii) (C) is correct
Justification:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
We know that, tan θ = sin θ/cos θ

sec θ = 1/ cos θ
cot θ = cos θ/sin θ
cosec θ = 1/sin θ

Now, substitute the above values in the given problem, we get
= (1 + sin θ/cos θ + 1/ cos θ) (1 + cos θ/sin θ – 1/sin θ)

Simplify the above equation,
= (cos θ +sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ
= (cos θ+sin θ)2-12/(cos θ sin θ)
= (cos2θ + sin2θ + 2cos θ sin θ -1)/(cos θ sin θ)
= (1+ 2cos θ sin θ -1)/(cos θ sin θ) (Since cos2θ + sin2θ = 1)
= (2cos θ sin θ)/(cos θ sin θ) = 2

Therefore, (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2

(iii) (D) is correct.
Justification:
We know that,

Sec A= 1/cos A
Tan A = sin A / cos A

Now, substitute the above values in the given problem, we get

(secA + tanA) (1 – sinA)
= (1/cos A + sin A/cos A) (1 – sinA)
= (1+sin A/cos A) (1 – sinA)
= (1 – sin2A)/cos A
= cos2A/cos A = cos A

Therefore, (secA + tanA) (1 – sinA) = cos A

(iv) (D) is correct.
Justification:
We know that,
tan2A =1/cot2A

Now, substitute this in the given problem, we get

1+tan2A/1+cot2A
= (1+1/cot2A)/1+cot2A
= (cot2A+1/cot2A)×(1/1+cot2A)
= 1/cot2A = tan2A

So, 1+tan2A/1+cot2A = tan2A

Question 5:
Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.

(i) (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin2A/(1-cos A)

[Hint : Simplify LHS and RHS separately]
(v) ( cos A–sin A+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec2A = 1+cot2A.
(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
(viii) (sin A + cosec A)+ (cos A + sec A)2 = 7+tan2A+cot2A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 =tan2A

Anwser 5:
(i) (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)
L.H.S. =  (cosec θ – cot θ)2
= (cosec2θ + cot2θ – 2cosec θ cot θ)
= (1/sin2θ + cos2θ/sin2θ – 2cos θ/sin2θ)
= (1 + cos2θ – 2cos θ)/(1 – cos2θ)
= (1-cos θ)2/(1 – cosθ)(1+cos θ)
= (1-cos θ)/(1+cos θ) = R.H.S.

(ii)  cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
= [cos2A + (1+sin A)2]/(1+sin A)cos A
= (cos2A + sin2A + 1 + 2sin A)/(1+sin A)cos A
= (1 + 1 + 2sin A)/(1+sin A)cos A
= (2+ 2sin A)/(1+sin A)cos A
= 2(1+sin A)/(1+sin A)cos A
= 2/cos A = 2 sec A = R.H.S.

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]
= [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]
= sin2θ/[cos θ(sin θ-cos θ)] + cos2θ/[sin θ(cos θ-sin θ)]
= sin2θ/[cos θ(sin θ-cos θ)] – cos2θ/[sin θ(sin θ-cos θ)]
= 1/(sin θ-cos θ) [(sin2θ/cos θ) – (cos2θ/sin θ)]
= 1/(sin θ-cos θ) × [(sin3θ – cos3θ)/sin θ cos θ]
= [(sin θ-cos θ)(sin2θ+cos2θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
= (1 + sin θ cos θ)/sin θ cos θ
= 1/sin θ cos θ + 1
= 1 + sec θ cosec θ = R.H.S.

(iv)  (1 + sec A)/sec A = sin2A/(1-cos A)
L.H.S. = (1 + sec A)/sec A
= (1 + 1/cos A)/1/cos A
= (cos A + 1)/cos A/1/cos A
= cos A + 1
R.H.S. = sin2A/(1-cos A)
= (1 – cos2A)/(1-cos A)
= (1-cos A)(1+cos A)/(1-cos A)
= cos A + 1
L.H.S. = R.H.S.

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
Dividing Numerator and Denominator by sin A,
= (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A
= (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)
= (cot A – cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2A – cot2A = 1)
= [(cot A + cosec A) – (cosec2A – cot2A)]/(cot A+ 1 – cosec A)
= [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)
=  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
=  cot A + cosec A = R.H.S.

(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
L.H.S. = (sin θ – 2sin3θ)/(2cos3θ – cos θ)
= [sin θ(1 – 2sin2θ)]/[cos θ(2cos2θ- 1)]
= sin θ[1 – 2(1-cos2θ)]/[cos θ(2cos2θ -1)]
= [sin θ(2cos2θ -1)]/[cos θ(2cos2θ -1)]
= tan θ = R.H.S.

(viii) (sin A + cosec A)+ (cos A + sec A)2 = 7+tan2A+cot2A
L.H.S. = (sin A + cosec A)+ (cos A + sec A)2
= (sin2A + cosec2A + 2 sin A cosec A) + (cos2A + sec2A + 2 cos A sec A)
= (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A
= 1 + 2 + 2 + 2 + tan2A + cot2A
= 7+tan2A+cot2A = R.H.S.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
= (1/sin A – sin A)(1/cos A – cos A)
= [(1-sin2A)/sin A][(1-cos2A)/cos A]
= (cos2A/sin A)×(sin2A/cos A)
= cos A sin A
R.H.S. = 1/(tan A+cotA)
= 1/(sin A/cos A +cos A/sin A)
= 1/[(sin2A+cos2A)/sin A cos A]
= cos A sin A
L.H.S. = R.H.S.

(x)  (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 =tan2A
L.H.S. = (1+tan2A/1+cot2A)
= (1+tan2A/1+1/tan2A)
= 1+tan2A/[(1+tan2A)/tan2A]
= tan2A