Class X Chapter 8 – Introduction to Trigonometry Maths

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Exercise 8.1
Question 1:
In ∆ABC right angled at B, AB = 24 cm, BC = 7 m. Determine
(i) sin A, cos A
(ii) sin C, cos C
Answer:
Applying Pythagoras theorem for ∆ABC, we obtain
AC
2 = AB2 + BC2= (24 cm)2 + (7 cm)2= (576 + 49) cm2= 625 cm2AC = cm = 25 cm
(i) sin A =
cos A =
(ii)

sin C =
cos C =
Question 2:In the given figure find tan P - cot R
Answer:
Applying Pythagoras theorem for ∆PQR, we obtain
PR
2 = PQ2 + QR2(13 cm)2 = (12 cm)2 + QR2
169 cm2 = 144 cm2 + QR225 cm2 = QR2QR = 5 cm
tan P - cot R =
Question 3:If sin A = , calculate cos A and tan A.
Answer:
Let ∆ABC be a right-angled triangle, right-angled at point B.

Given that,
Let BC be 3
k. Therefore, AC will be 4k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC, we obtain
AC
2 = AB2 + BC2(4k)2 = AB2 + (3k)216k 2 - 9k 2 = AB27k 2 = AB2AB =Question 4:
Given 15 cot A = 8. Find sin A and sec A
Answer:
Consider a right-angled triangle, right-angled at B.
It is given that,
cot A =
Let AB be 8
k.Therefore, BC will be 15k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC, we obtain
AC
2 = AB2 + BC2= (8k)2 + (15k)2= 64k2 + 225k2= 289k2AC = 17k
Question 5:Given sec θ = , calculate all other trigonometric ratios.
Answer:
Consider a right-angle triangle ∆ABC, right-angled at point B.
If AC is 13
k, AB will be 12k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC, we obtain
(AC)
2 = (AB)2 + (BC)2(13k)2 = (12k)2 + (BC)2169k2 = 144k2 + BC225k2 = BC2
BC = 5kQuestion 6:If A and B are acute angles such that cos A = cos B, then show thatA = B.
Answer:
Let us consider a triangle ABC in which CD
AB.
It is given that
cos A = cos B
… (1)

We have to prove A = B. To prove this, let us extend AC to P such that BC = CP.
From equation (1), we obtain
By using the converse of B.P.T,
CD||BP
⇒∠ ACD = CPB (Corresponding angles) … (3)
And,
BCD = CBP (Alternate interior angles) … (4)
By construction, we have BC = CP.
∴ ∠ CBP = CPB (Angle opposite to equal sides of a triangle) … (5)
From equations (3), (4), and (5), we obtain
ACD = BCD … (6)
In ∆CAD and ∆CBD,
ACD = BCD [Using equation (6)]CDA = CDB [Both 90°]
Therefore, the remaining angles should be equal.
∴∠ CAD = CBD⇒ ∠ A = BAlternatively,
Let us consider a triangle ABC in which CD AB.
It is given that,
cos A = cos B
Let
AD = k BD … (1)
And, AC =
k BC … (2)
Using Pythagoras theorem for triangles CAD and CBD, we obtain
CD
2 = AC2 - AD2 … (3)
And, CD
2 = BC2 - BD2 … (4)
From equations (3) and (4), we obtain
AC
2 - AD2 = BC2 - BD2(k BC)2 - (k BD)2 = BC2 - BD2k2 (BC2 - BD2) = BC2 - BD2k2 = 1k = 1
Putting this value in equation (2), we obtain
AC = BC
⇒ ∠ A = B(Angles opposite to equal sides of a triangle)
Question 7:If cot θ = , evaluate
(i) (ii) cot
2 θ
Answer:
Let us consider a right triangle ABC, right-angled at point B.
If BC is 7
k, then AB will be 8k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC, we obtain
AC
2 = AB2 + BC2= (8k)2 + (7k)2= 64k2 + 49k2= 113k2AC =
(i)
(ii) cot
2 θ = (cot θ)2 = =Question 8:If 3 cot A = 4, Check whether
Answer:
It is given that 3cot A = 4
Or, cot A =

Consider a right triangle ABC, right-angled at point B.
If AB is 4
k, then BC will be 3k, where k is a positive integer.
In ∆ABC,
(AC)
2 = (AB)2 + (BC)2= (4k)2 + (3k)2= 16k2 + 9k2= 25k2AC = 5k
cos2 A - sin2 A =Question 9:In ∆ABC, right angled at B. If , find the value of
(i) sin A cos C + cos A sin C
(ii) cos A cos C - sin A sin C
Answer:

If BC is k, then AB will be , where k is a positive integer.
In ∆ABC,
AC
2 = AB2 + BC2=
= 3
k2 + k2 = 4k2AC = 2k(i) sin A cos C + cos A sin C
(ii) cos A cos C - sin A sin C

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.
Answer:
Given that, PR + QR = 25
PQ = 5
Let PR be
x.
Therefore, QR = 25 -
xApplying Pythagoras theorem in ∆PQR, we obtain
PR
2 = PQ2 + QR2x2 = (5)2 + (25 - x)2x2 = 25 + 625 + x2 - 50x50x = 650x = 13
Therefore, PR = 13 cm
QR = (25 - 13) cm = 12 cm

Question 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 = 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 = , for some angle θ
Answer:
(i) Consider a ∆ABC, right-angled at B.

But > 1tan A > 1
So, tan A < 1 is not always true.
Hence, the given statement is false.
(ii)
Let AC be 12
k, AB will be 5k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC, we obtain
AC
2 = AB2 + BC2(12k)2 = (5k)2 + BC2144k2 = 25k2 + BC2BC2 = 119k2BC = 10.9kIt can be observed that for given two sides AC = 12k and AB = 5k,
BC should be such that,
AC - AB < BC < AC + AB
12
k - 5k < BC < 12k + 5k7k < BC < 17 k
However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of
sec A is possible.
Hence, the given statement is true.
(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the
abbreviation used for cosine of angle A.
Hence, the given statement is false.
(iv) cot A is not the product of cot and A. It is the cotangent of
A.
Hence, the given statement is false.
(v) sin θ =
We know that in a right-angled triangle,
In a right-angled triangle, hypotenuse is always greater than the remaining two
sides. Therefore, such value of sin θ is not possible.
Hence, the given statement is false

Exercise 8.2
Question 1:
Evaluate the following
(i) sin60° cos30° + sin30° cos 60°
(ii) 2tan
245° + cos230° - sin260°
(iii)
(iv)
(v)
Answer:
(i) sin60° cos30° + sin30° cos 60°
(ii) 2tan
245° + cos230° - sin260°
(iii)

(iv)
(v)
Question 2:Choose the correct option and justify your choice.
(i)
(A). sin60°
(B). cos60°
(C). tan60°
(D). sin30°
(ii)
(A). tan90°
(B). 1
(C). sin45°
(D). 0
(iii) sin2A = 2sinA is true when A =
(A). 0°
(B). 30°
(C). 45°
(D). 60°
(iv)
(A). cos60°
(B). sin60°
(C). tan60°
(D). sin30°
Answer:

(i)
Out of the given alternatives, only
Hence, (A) is correct.
(ii)
Hence, (D) is correct.
(iii)Out of the given alternatives, only A = 0° is correct.
As sin 2A = sin 0° = 0
2 sinA = 2sin 0° = 2(0) = 0
Hence, (A) is correct.
(iv)

Out of the given alternatives, only tan 60°
Hence, (C) is correct.
Question 3:If and ;
0° < A + B ≤ 90°, A > B find A and B.
Answer:
⇒ ⇒A + B = 60 … (1)tan (A - B) = tan30A - B = 30 … (2)
On adding both equations, we obtain
2A = 90
A = 45
From equation (1), we obtain
45 + B = 60
B = 15
Therefore,
A = 45° and B = 15°Question 4:State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B

(ii) The value of sinθ increases as θ increases
(iii) The value of cos θ increases as θ increases
(iv) sinθ = cos θ for all values of θ
(v) cot A is not defined for A = 0°
Answer:
(i) sin (A + B) = sin A + sin B
Let A = 30° and B = 60°
sin (A + B) = sin (30° + 60°)
= sin 90°
= 1
sin A + sin B = sin 30° + sin 60°
Clearly, sin (A + B) ≠ sin A + sin B
Hence, the given statement is false.
(ii) The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as
sin 0° = 0
sin 90° = 1
Hence, the given statement is true.
(iii) cos 0° = 1

cos90° = 0
It can be observed that the value of cos θ does not increase in the interval of 0° < θ
< 90°.
Hence, the given statement is false.
(iv) sin θ = cos θ for all values of θ.
This is true when θ = 45°
As
It is not true for all other values of θ.
As and ,
Hence, the given statement is false.
(v) cot A is not defined for A = 0°
As ,
= undefined
Hence, the given statement is true.

Exercise 8.3
Question 1:
Evaluate
(I)
(II)
(III) cos 48° - sin 42°
(IV)cosec 31° - sec 59°
Answer:
(I)
(II)
(III)cos 48° - sin 42° = cos (90°- 42°) - sin 42°
= sin 42° - sin 42°
= 0
(IV) cosec 31° - sec 59° = cosec (90° - 59°) - sec 59°
= 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

Answer:
(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.
Answer:
Given that,
tan 2A = cot (A- 18°)
cot (90° - 2A) = cot (A -18°)
90° - 2A = A- 18°
108° = 3A
A = 36°
Question 4:If tan A = cot B, prove that A + B = 90°
Answer:
Given that,
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.
Answer:
Given that,
sec 4A = cosec (A - 20°)
cosec (90° - 4A) = cosec (A - 20°)
90° - 4A= A- 20°
110° = 5A
A = 22°
Question 6:If A, Band C are interior angles of a triangle ABC then show that
Answer:
We know that for a triangle ABC,
A + B + C = 180°B + C= 180° - AQuestion 7:Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and
45°.
Answer:
sin 67° + cos 75°

= sin (90° - 23°) + cos (90° - 15°)= cos 23° + sin 15°
Exercise 8.4
Question 1:
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Answer:
We know that,
will always be positive as we are adding two positive quantities.
Therefore,
We know that,
However,
Therefore,
Also,

Question 2:Write all the other trigonometric ratios of A in terms of sec A.
Answer:
We know that,
Also, sin
2 A + cos2 A = 1
sin
2 A = 1 - cos2 A
tan
2A + 1 = sec2A
tan
2A = sec2A - 1Question 3:Evaluate
(i)

(ii) sin25° cos65° + cos25° sin65°
Answer:
(i)
(As sin
2A + cos2A = 1)
= 1
(ii) sin25° cos65° + cos25° sin65°
= sin
225° + cos225°
= 1 (As sin
2A + cos2A = 1)Question 4:Choose the correct option. Justify your choice.
(i) 9 sec
2 A - 9 tan2 A =
(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)
(A) sec
2 A
(B) -1
(C) cot
2 A
(D) tan
2 A
Answer:
(i) 9 sec
2A - 9 tan2A
= 9 (sec
2A - tan2A)
= 9 (1) [As sec
2 A - tan2 A = 1]
= 9
Hence, alternative (B) is correct.
(ii)
(1 + tan θ + sec θ) (1 + cot θ - cosec θ)

Hence, alternative (C) is correct.
(iii) (secA + tanA) (1 - sinA)
= cosA
Hence, alternative (D) is correct.
(iv)

Hence, alternative (D) is correct.Question 5:Prove the following identities, where the angles involved are acute angles for which
the expressions are defined.
Answer:
(i)
(ii)

(iii)
= secθ cosec θ +
= R.H.S.
(iv)

= R.H.S
(v)
Using the identity cosec
2 = 1 + cot2 ,
L.H.S =

= cosec A + cot A
= R.H.S
(vi)

(vii)
(viii)

(ix)
Hence, L.H.S = R.H.S
(x)

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