# Introduction to Trigonometry - NCERT Questions

Q 1.

In ΔABC , right-angled at B , AB = 24 cm,
BC = 7 cm. Determine :
(i) sin A , cos A (ii) sin C , cos C

SOLUTION:

Soln. : In right angle ΔABC, we have AB = 24 cm, BC = 7 cm

∴ Using Pythagoras theorem, AC2 = AB2 + BC2
AC2 = 242 + 72 = 576 +49 = 625 = 252
AC = 25 cm
(i)
(ii) ,

Q 2.

In the figure, find tan P - cot R .

SOLUTION:

Soln. : In right angle ΔPQR
Using the Pythagoras theorem, we get
QR2 = PR2 - PQ2
QR2 = 132 - 122 = (13 - 12)(13 + 12)
= 1 × 25 = 25
cm
Now, ,

Q 3.

If calculate cos A and tan A .

SOLUTION:

Soln. : Let us consider, the right ΔABC, we have
Perpendicular = BC and

Hypotenuse = AC

Let BC = 3k and AC = 4k

= (4k)2 - (3k)2 = (4k - 3k)(4k + 3k) = k(7k) = 7k2

Q 4.

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

SOLUTION:

Soln. : In the right triangle ABC, we have
15 cot A = 8

AB = 8k and BC = 15k
Now, using Pythagoras theorem, we get
AC2 = AB2 + BC2
= (8k)2 + (15k)2 = 64k2 + 225k2 = 289k2 = (17k)2
AC =

Q 5.

Given sec θ = calculate all other trigonometric ratios.

SOLUTION:

Soln. : In right ΔABC, ∠ B = 90°

Let ∠ A = θ and sec θ =

AC = 13k and AB = 12k
∴ Using Pythagoras theorem, we get
BC2 = AC2 - AB2
BC2 = (13k)2 - (12k)2
= (13k - 12k) (13k + 12k)
= k(25k) = 25k2 = (5k)2

Q 6.

If ∠ A and ∠ B are acute angles such that cos A = cos B , then show that ∠ A = ∠ B .

SOLUTION:

Soln. : Let us consider a right ΔABC, ∠ C = 90°
Now, cos A =

And
Since, cos A = cos B

Now, in the ΔABC, two sides AC and BC are equal. [Proved above]
∴ Their opposite angles are also equal.
∴ ∠ A = ∠ B.

Q 7.

If cot θ = evaluate:
(i) (ii) cot 2θ

SOLUTION:

Soln. : In right ΔABC, ∠ B = 90° and ∠ A = θ
.......(1) [Given]
But in right ΔABC, cot θ = .......(2)

From (1) and (2),
AB = 7k and BC = 8k
AC2 = AB2 + BC2 = (7k)2 + (8k)2
(Pythagoras theorem)

Now,
(i)

(ii)

Q 8.

If 3 cot A = 4, check whether

SOLUTION:

Soln. : In right angled ΔABC, ∠ B = 90°
∴ For ∠ A, we have:
Base = AB and perpendicular = BC. Also
Hypotenuse = AC

∴ 3 cot A = 4
∴ cot A = 4/3 ...(1)
But cot A ...(2)
∴ From (1) and (2),
AB = 4k and BC = 3k
∴ Using Pythagoras theorem, we get
AC2 = AB2 + BC2
AC2 = (4k)2 + (3k)2

Now,

Now, to check the given equation,
L.H.S. =
........(i)
R.H.S. = cos2A - sin2A
.........(ii)
From (i) and (ii), we have L.H.S. = R.H.S.

Q 9.

In triangle ABC , right angled at B , if tan A =
find the value of :

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

SOLUTION:

Soln. : In right ΔABC, ∠ B = 90°
For ∠ A, we have
Base = BC, Perpendicular = AB,
Hypotenuse = AC
.......(1) (Given)
∴ tan A ........(2)
From (1) and (2), we have

AB = and BC = 1k
Now, using Pythagoras theorem, we get
AC2 = AB2 + BC2

Now, ,

Again, for ∠ C, we have
Base = BC, Perpendicular = AB
and Hypotenuse = AC
,

(i)

(ii)

Q 10.

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

SOLUTION:

Soln. : In right ΔPQR, Q = 90°

PR + QR = 25 cm and PQ = 5 cm
Let QR = x cm ⇒ PR = (25 - x)
∴ By Pythagoras theorem, we have
PR2 = QR2 + PQ2
⇒ (25 - x)2 = x2 + 52
⇒ 625 - 50x + x2 = x2 +25
⇒ - 50x = - 600
i.e., RQ = 12 cm
RP = 25 - 12 = 13 cm
Now, , ,

Q 11.

(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) for some angle θ.

SOLUTION:

Soln. : (i) False [∵ A tangent of an angle is ratio of sides other than hypotenuse, which may be equal or unequal to each other.]
(ii) True [∵ cos A is always less than 1]
i.e., sec A will always be greater
than 1
(iii) False [∵ 'cosine A' is abbreviated as 'cos A']
(iv) False ['cot A' is a single and meaningful term whereas 'cot' alone has no meaning.]
(v) False [ is greater than 1 and sinθ cannot be greater than 1]

Q 12.

Evaluate the following:
(i) sin60° cos30° + sin 30° cos 60°
(ii) 2tan 2 45° + cos 2 30° - sin 2 60°
(iii)
(iv)
(v)

SOLUTION:

Soln. : (i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2tan2 45° + cos2 30° - sin2 60°

(iii)

(iv)

and cot 45° = 1

(v)

Q 13.

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

SOLUTION:

Soln. : (i) (A)

(ii) (D) :

(iii) (A) : When A = 0, then
sin 2A = sin 2(0°) = sin 0° = 0,
2 sin A = 2 sin 0° = 2 × 0 = 0
i.e., sin 2A = 2sin A for A = 0°
(iv) (C) :

Q 14.

If tan ( A + B ) = and tan ( A - B ) =
0° < A + B < 90°; A > B , find A and B.

SOLUTION:

Soln. : We have,
tan 60° , tan 30° = ......(1)
Also tan (A + B) and tan (A - B) (Given)
.......(2)
From (1) and (2), we get A + B = 60° and
A - B = 30°
On adding A + B and A - B, we get 2A = 90°
A = 45°
On subtracting A + B and A - B, we get 2B = 30°
B = 15°

Q 15.

(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°.

SOLUTION:

Soln. : (i) False : Let us take A = 30° and B = 60°, then L.H.S = sin (30° + 60°) = sin 90° = 1
R.H.S.= sin 30° + sin 60°

∴ L.H.S. ≠ R.H.S.
(ii) True : Since the values of sin θ increases from 0 to 1 as θ increases from 0 to 90°.
(iii) False : Since the value of cos θ decreases from 1 to 0 as θ increases from 0 to 90°.
(iv) False : Let us take θ = 30°

⇒ sin 30° ≠ cos 30°
(v) True : We have cot 0° = not defined

Q 16.

Evaluate:
(i) (ii)
(iii) (iv)

SOLUTION:

Soln. : (i)
⇒ sin 18° = sin (90° - 72°) = cos 72°
[∵ sin (90° - A) = cos A]

(ii)
tan 26° = tan (90° - 64°) = cot 64° [∵ tan (90° - A) = cot A]

(iii) cos 48° - sin 42°
cos 48° = cos (90° - 42°) = sin 42° [∵ cos (90° - A) = sin A]
∴ cos 48° - sin 42° = sin 42° - sin 42° = 0
(iv) cosec 31° - sec 59°
cosec 31° = cosec (90° - 59°) = sec 59° [∵ cosec (90° - A) = sec A]
∴ cosec 31° - sec 59° = sec 59° - sec 59° = 0

Q 17.

Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° - sin 38° sin 52° = 0

SOLUTION:

Soln. : (i) tan 48° tan 23° tan 42° tan 67° = 1
L.H.S. = tan 48° tan 23° tan 42° tan 67°
= tan (90° - 42°) tan 23° tan 42° tan (90° - 23°)
[Q tan (90° - A) = cot A]
= cot 42° tan 23° tan 42° cot 23°

∴ L.H.S. = R.H.S.
(ii) cos 38° cos 52° - sin 38° sin 52° = 0
L.H.S. = cos 38° cos 52° - sin 38° sin 52°
= cos 38° cos (90° - 38°) - sin 38° sin (90° - 38°)
= cos 38° sin 38° - sin 38° cos 38°
[∵ sin (90° - A) = cos A and cos (90° - A)= sin A]
= 0 = R.H.S.
∴ L.H.S. = R.H.S.

Q 18.

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

SOLUTION:

Soln. : Since tan 2A = cot (A - 18°)
Also tan (2A) = cot (90° - 2A)
[∵ tan θ = cot (90° - θ)]
A - 18° = 90° - 2A
A + 2A = 90° + 18°
⇒ 3A = 108° ⇒

Q 19.

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

SOLUTION:

Soln. : tan A = cot B and cot B = tan (90° - B)
A = 90° - BA + B = 90°
[∵ tan (90° - θ) = cot θ]

Q 20.

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

SOLUTION:

Soln. : ∵ sec 4A = cosec (A - 20°)
sec (4A) = cosec (90° - 4A)
[∵ cosec (90° - θ) = sec θ]
A - 20° = 90° - 4A
A + 4A = 90° + 20°
⇒ 5A = 110° ⇒ A =

Q 21.

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

SOLUTION:

Soln. : Since, sum of the angles of ΔABC is
A + B + C = 180°
B + C = 180° - A
Dividing both sides by 2, we get

[∵ sin (90° - θ) = cos θ]

Q 22.

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

SOLUTION:

Soln. : Since sin 67° = sin (90° - 23°) = cos 23°
[∵ sin (90° - θ ) = cos θ]
Also, cos 75° = cos (90° - 15°) = sin 15°
[∵ cos (90° - θ) = sin θ]
∴ sin 67° + cos 75° = cos 23° + sin 15°

Q 23.

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

SOLUTION:

Soln. : (a) sin A

(b)

(c)

Q 24.

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

SOLUTION:

Soln. : (i) sin A =

(ii)
(iii)
(iv)

(v)

Q 25.

Evaluate :
(i)
(ii)

SOLUTION:

Soln. : (i)
∵ sin 63° = sin (90° - 27°) = cos 27°
⇒ sin2 63° = cos2 27°
cos2 73° = cos2 (90° - 17°) = sin2 17°
= 1
[∵ cos2A + sin2A = 1]
(ii) sin 25° cos 65° + cos 25° sin 65°
∵ sin 25° = sin (90° - 65°) = cos 65°
[∵ sin (90° - A) = cos A]
cos 25° = cos (90° - 65°) = sin 65°
[∵ cos (90° - A) = sin A]
∴ sin 25° cos 65° + cos 25° sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= (cos 65°)2 + (sin 65°)2
= cos2 65° + sin2 65° = 1 [∵ cos2 A + sin2 A = 1]

Q 26.

Choose the correct option. Justify your choice.
(i) 9 sec 2 A - 9 tan 2 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) (sec A + tan A ) (1 - sin A ) = .......................
(A) sec A (B) sin A
(C) cosec A (D) cos A
(iv) .........................
(A) sec 2 A (B) - 1
(C) cot 2 A (D) tan 2 A

SOLUTION:

Soln. : (i) (B) : Since, 9 sec2 A - 9 tan2 A
= 9 (sec2 A - tan2 A) = 9 (1) = 9
[∵ tan2 A + 1 = sec2 A ⇒ sec2 A - tan2A = 1]
(ii) (C) : Here, (1 + tan θ + sec θ) (1 + cot θ - cosec θ)
= (1 + tan θ + sec θ)
= (1 + tan θ + sec θ)

= 2
(iii) (D) : We have : (sec A + tan A) (1 - sin A)

= cos A
(iv) (D) : Here,
= tan2 A

Q 27.

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ - cot θ ) 2 =
(ii)
(iii)
[Hint : Write the expression in terms of sin θ and cos θ ]
(iv) [Hint : Simplify L.H.S. and R.H.S. separately]
(v) = cosec A + cot A , using the identity cosec 2 A = 1 + cot 2 A .
(vi)
(vii)
(viii) (sin A + cosec A ) 2 + (cos A + sec A ) 2
= 7 + tan 2 A + cot 2 A
(ix) (cosec A - sin A ) (sec A - cos A )

[Hint : Simplify L.H.S. and R.H.S. separately]
(x)

SOLUTION:

Soln. : (i) L.H.S = (cosec θ - cot θ)2

[∵ sin2 θ = 1 - cos2 θ]

[∵ 1 - cos2 θ = (1 - cos θ) (1 + cos θ)]

Hence, L.H.S. = R.H.S.
(ii) L.H.S.

[∵ cos2 A + sin2 A = 1]

= R.H.S
Hence, L.H.S. = R.H.S.
(iii) L.H.S.

[∵ sin2 θ + cos2 θ = 1]

= cosec θ sec θ + 1 = 1 + sec θ·cosec θ
= R.H.S.
(iv)

[Multiplying and dividing by (1 - cos A)]

= R.H.S. [∵ 1 - cos2 A = sin2 A]
(v) L.H.S.
[Dividing each term of num. and den. by sin A]

[Multiplying and dividing by (cot A + cosec A)]

[∵ cot2 A - cosec2 A = - 1]
= cot A + cosec A = R.H.S.
(vi) L.H.S.
[Multiplying and dividing by (cot A + cosec A)]
[∵ (1 - sin A) (1 + sin A)
= 1 - sin2 A]
[∵ 1 - sin2 A = cos2 A]

= sec A + tan A = R.H.S.
(vii)

(viii) L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
= sin2 A + cosec2 A + 2 sin A·cosec A + cos2 A
+ sec2 A + 2 cos A·sec A
= (sin2 A + cos2 A) + cosec2 A + sec2 A + 2 + 2
[sin A·cosec A = 1 and sec A·cos A = 1]
= 1 + cosec2 A + sec2 A + 4
[∵ sin2 A + cos2 A = 1]
= 5 + (1 + cot2 A) + (1 + tan2 A)
[∵ cosec2 A = 1 + cot2 A and sec2 A = 1 + tan2 A]
= 7 + cot2 A + tan2 A = R.H.S
(ix) L.H.S. = (cosec A - sin A) (sec A - cos A)

[∵ 1 - sin2 A = cos2 A
and 1 - cos2 A = sin2 A]
= sin A·cos A

[∵ 1 = sin2 A + cos2 A]
[Dividing num. and den. by sin A cos A]

(x)

= tan2 A = R.H.S. ...(1)

= (-tan A)2 = tan2A = R.H.S. ...(2)
∴ From (1) and (2), we get