# Trigonometrical Identities Concise Solutions Chapter-21 Class 10

Trigonometrical Identities Concise Solutions Chapter-21 Class 10. Selina  Concise Solutions of Exercise – 21 (A), Exercise – 21 (B), Exercise –21 (C), Exercise – 21 (D), Exercise –21 (E) for Class 10th. Concise Maths Solutions Trigonometrical Identities Chapter-21 for ICSE Maths Class 10 is available here. All Solutions of Concise Selina Maths of Trigonometrical Identities Chapter-21 has been solved according instruction given by council. This Post is the  Concise Solutions of Trigonometrical Identities Chapter-21 for ICSE Maths Class 10th.

Trigonometrical Identities Concise Solutions Chapter-21 Class 10

The Solutions of Concise Mathematics Trigonometrical Identities  Chapter-21 for ICSE Class 10. Experience teachers Solved Chapter-21 Trigonometrical Identities  of  Concise Selina Maths  for class 10th ICSE board. Therefore the ICSE Class 10th Maths Solutions of Concise Selina Publishers is helpful on  various topics including Chapter-21 Trigonometrical Identities.

–:Select Topic :–

Exercise – 21 (A),

Exercise – 21 (B),

Exercise –21 (C)

Exercise – 21 (D),

Exercise – 21 (E),

### EXERCISE -21(A) Trigonometrical Identities Concise Selina Maths Solutions for ICSE Class 10

Prove the following Identities :

…………

………..

……….

………..

#### Question 5.

sin4A – cos4 A = 2 sin2A-1

L.H.S. = sin4 A – cos4A = (sin2A)2-(cos2A)2
= (sin2A + cos2A) (sin2A – cos2A)     [(a2 – b2 = (a + b) (a – b)]
and = 1 (sin2 A – cos2A) [∵ sin2A + cos2A = 1]
so = sin2 A – (1- sin2A) (∵ cos2A = 1 – sin2A)
therefore = sin2 A – 1 + sin2 A
hence = 2 sin2A-1 = R.H.S.

#### Question 6.

(1 – tan A)2 + (1 + tanA)2 = 2sec2A

LHS = (1 -tanA)2 + (1 +tanA)2
= 1 + tan2 A- 2 tan A + 1 + tan2 A + 2 tanA
and = 2 + 2 tan2 A = 2 (1+tan2A)
hence = 2 sec2A (∵ l+tan2A=sec2A)
= R.H.S.

#### Question 7.

Cosec4 A – cosec2 A = cot4 A + cot2 A

L.H.S. = cosec4 A -cosec2 A
= (cosec2A)2 – cosec2A
and = (1 + cot2A)2 – (1 + cot2A)
= 1 + cot4 A + 2 cot2A – 1- cot2A
hence = cot4 A + cot2 A = R.H.S.

#### Question 8.

sec A (1-sin A) (sec A + tan A) = 1

#### Question 9.

cosec A (1 + cos A) (cosec A – cot A) = 1

#### Question 10.

sec2 A + cosec2A = sec2 A cosec2 A

…………

#### Question 12.

tan2A – sin2A = tan2 A. sin2 A

#### Question 13.

cot2 A – cos2 A = cos2 A. cot2 A

#### Question 14.

(cosecA + sinA) (cosec A – sinA) = cot2 A + cos2A

L.H.S. = (cosec A + sin A) (cosec A – sin A)
= (cosec2A – sin2 A) [∵ (a + b) (a – b) = a2– b2]
and = 1 + cot2 A – sin2 A = cot2 A + 1 – sin2A
= cot2 A + cos2 A (∵ 1-sin2A = cos2 A)
hence = R.H.S.

#### Question 15

(sec A – cosA) (sec A + cosA) = sin2 A + tan2A

L.H.S. = (sec A-cos A) (sec A + cos A)
= sec2 A – cos2 A
and = 1 + tan2A-cos2 A
= 1-cos2 A + tan2 A
so = sin2 A + tan2 A  (∵ 1- cos2A=sin2A)
hence = R.H.S.

#### Question 16.

(cos A + sin A)2 + (cos A – sin A)2 = 2

LHS = (cos A + sin A)2 + (cos A – sin A)2
= cos2 A + sin2 A + 2 cos A sin A + cos2 A + sin2 A – 2 cos A sin A
and = 2 sin2 A + 2 cos2 A
= 2 (sin2A+cos2A)
hence = 2 x 1=2 = R.H.S. (∵ sin2A + cos2 A = 1)

#### Question 17.

(cosec A – sinA) (sec A – cos A) (tan A + cot A) = 1

……

……….

………..

#### Question 21.

…………..
(sin A + cosec A)2 + (cos A + sec A)= 7 + tan2 A + cot2 A

L.H.S. = (sin A + cosecA)2 + (cosA+ secA)2
= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A
and = sin2 A+cosec2 A+2 sin A x $\frac { 1 }{ sinA }$ + cos2 A+sec2A + 2cosA x $\frac { 1 }{ cosA }$
= sin2A + cos2 A + cosec2 A + sec2A+ 2 + 2   (∵ sin2 A + cos2A= 1)
so = 1 +cosec2A + sec2A + 4
= (1 + cot2 A) + (1 + tan2 A) + 5 [∵ cosec2A = 1 + cot2 A and sec2 A = 1 + tan2A]
hence = 1 + cot2 A + 1 + tan2 A + 5
= 7 + tan2A + cot2A = R.H.S.

#### Question 22.

sec2A. cosec2A = tan2A + cot2A + 2

……..

………..

…………

………….

………….

…………

…………..

…………..

……………

………….

……………

……………

………………

…………..

………….

#### Question 38.

(1 +cot A-cosec A) (1 + tan A + sec A) = 2

………..

……………

…………..

………….

………….

…………

…………

…………..

……………

…………..

### EXERCISE – 21 (B) Solutions of Selina Concise Mathematics Chapter-20 Trigonometrical Identities

Prove that:

(i)…

(ii)…….

(iii)………

(iv)………

(v)………

(vi)……..

(vii)…….

(viii)…….

(ix)……….

#### Question 2.

If xcosA + ysinA = m and xsinA-ycosA = n, then prove that: x2+y2 = m2 + n2

x cos A + y sin A = m    …(i)
x sin A – y cos A = n     ….(ii)
squaring (i) and (ii)
x2 cos2 A + y2 sin2 A + 2 xy cosA sinA = m2
x2 sin2 A + y2 cos2 A – 2 xy cos A sin A = n2
x2 (sin2 A + cos2 A) + y2 (sin2 A + cos2 A) = m2+n2
∴ x2+y2 = m2 + n2(∵ sin2A + cos2A= 1)
Hence proved.

#### Question 3.

If m = a sec A +b tan A and n=atanA + bsecA, then prove that: m2-n2 = a2-b2

m = asec A + btan A         ……(i)
n = a tan A + b sec A       …..(ii)
squaring (i) and (ii)
m2 = a2 sec2 A + b2 tan2 A + 2ab sec A tan A
n2 = a2 tan2 A + b2 sec2 A + 2 ab tan A sec A
Subtracting, we get
m2 – n2 = a2 (sec2 A – tan2 A) + b2 (tan2 A – sec2 A)
= a2x 1 +b2(-1) = a2-b( ∵ sec2A-tan2A= 1)  .
Henceproved

#### Question 4.

If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that: x2 + y2 + z2 = r.

x = r sin A cos B      ….(i)
y = r sin A sin B      ….(ii)
z = r cosA               …….(iii)
Squaring, (i), (ii) & (iii)
x2=r2 sin2 A cos2 B,
y2 = r2sin2Asin2B,
z2 = r2cos2A
x2+y2 + z2=r2 (sin2A cos2E + sin2 A sin2 B+cos2A)
= r[sin2A (cos2 B + sin2B) + cos2A]
and = r [sin2 A x 1 + cos2 A]
so = r2 [sin2 A + cos2 A] = r2 x 1  = r2        ( ∵ sin2 A + cos2 A = 1)
Hence proved.

#### Question 5.

If sin A + cos A = m and sec A + cosec A=n, show that n (m2-1) = 2m

#### Question 6.

If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that x2 + y2 + z2 = r2

x = r cosAcosB              ….(i)
y = r cosAsinB             ….(ii)
z = r sinA                 ….(iii)
Squaring (i), (ii), (iii)
x2 = r2 cos2 A cos2 B, y2 = r2 cos2 A sin2B
z2 = r2sin2A
x2 + y2 + z2 = r2 (cos2 A cos2B + cos2 A sin2 B + sin2 A)
= r2 [cos2 A (cos2 B + sin2B) + sin2 A]
= r2[cos2Ax 1+sin2A]
= r2 (1) = r2    `Hence proved.

### Chapter-21 Trigonometrical Identities Concise Solutions EXERCISE – 21 (C)

#### Question 1.

Show that:
(i) tan 10° tan 15° tan 75° tan 80° = 1
(ii) sin 42° sec 48°+cos 42° cosec 48°= 2
(iii) ………

(i)

tan 10° tan 15° tan 75° tan 80°= 1
L.H.S. = tan 10° tan 15° tan 75° tan 80°
= tan (90° – 80°) tan (90° – 75°) tan 75° tan 80°
and = cot 80° cot 75° tan 75° tan 80°
so = tan 80° cot 80° x tan 75° cot 75°
hence = 1 x 1 = 1= R.H.S. (∵ tan A cot A = 1)
(ii)

sin 42° sec 48°+ cos 42° cosec 48°= 2
L.H.S. = sin 42° sec 48°+ cos 42° cosec 48°
= sin 42° sec (90° – 42°) + cos 42° cosec (90° – 42°)
= sin 42° cosec 42°+ cos 42° sec 42°
=1 + 1=2 R.H.S. (∵ sin A cosec A=1, cos A sec A=1)

#### Question 2.

Express each of the following in terms of angles between 0°and 45°.
(i) sin 59° + tan 63°
(ii) cosec 68° + cot 72°
(iii) cos 74° + sec 67°

(i)

sin 59° + tan 63°
= sin (90° – 31°) + tan (90° – 27°)
= cos 31°+ cot 27°
(ii)

cosec 68° + cot 72°
= cosec (90° – 22°) + cot (90° – 18°)
= sec 22°+ tan 18°
(iii)

cos 74°+ sec 67°
= cos (90° – 16°) + sec (90° – 23°)
= sin 16°+ cosec 23°

Show that:
…………

#### Question 4.

For triangle ABC, Show that:

Evaluate:

(i)…

(ii)…….

(iii)………

(iv)………

(v)………

(vi)……..

(vii)…….

(viii)…….

(ix)……….

(i)

………..

#### Question 7.

Find (in each case, given below) the value of x, if:
(i) sin x = sin 60° cos 30° – cos 60° sin 30°
(ii) sin x = sin 60° cos 30° + cos 60° sin 30°
(iii) cos x = cos 60° cos 30° – sin 60° sin 30°
(iv) ………….
(v) sin 2x = 2 sin 45° cos 45°
(vi) sin 3x = 2 sin 30° cos 30°
(vii) cos (2x – 6°) = cos2 30° – cos2 60°

#### Question 8.

In each case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
(i) sin (90° – 3A). cosec 42° = 1
(ii) cos (90° – A). sec 77° = 1

Prove that:

…………….

Evaluate:

………….

#### Question 11.

Without using trigonometric tables, evaluate sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot30°.

#### Question 12.

Without using trigonometrical tables, evaluate: cosec2 57° – tan2 33° + cos 44° cosec 46° – $\sqrt{2}$ cos45°- tan2 60°

### Selina Publishers Concise Maths Solutions of EXERCISE – 21 (D)Trigonometrical Identities

#### Question 1.

Use tables to find sine of:
(i) 21°
(ii) 34°42′
(iii) 47° 32′
(iv) 62°57′
(v) 10°20′ + 20° 45′

From tables of sine of angles, we find that:
(i)

sin 21°= 0.3584,
(ii)

sin 34°42’= .5693
(iii)

sin 47° 32′ = 0.7377
(iv)

sin 62° 57′ = 0.8906
(v)

sin 10° 20′ + 20°45′ = sin 31°5′
= 0.5162

#### Question 2.

Use tables to find cosine of:
(i) 2°4′
(ii) 8°12′
(iii) 26°32’
(iv) 65°41′
(v) 9°23′ +15°54′

From tables of cosine of angle, we find that:
(i)

cos 2°4′ = 0.9993
(ii)

cos 8° 12’ = 0.9898
(iii)

cos 26°32′ = 0.8946
(iv)

cos 65°41′ = 0.4118
(v)

cos 9°23′ + 15°54′ = cos 25° 17′
= 0.9042

#### Question 3.

Use trigonometrical tables to find tangent of:
(i) 37°
(ii) 42°18′
(iii) 17°27′

From the tables of tangents, we find that
(i)

tan 35° = 0.7536
(ii)

tan 42°18’= 0.9099
(iii)

tan 17°27’= 0.3144

#### Question 4.

Use tables to find the acute angle θ, if the value of sin θ
(i) 4848
(ii) 0.3827
(iii) 0.6525

From the tables of series, we find that of :
(i)

sinθ = 0.4848, then θ = 29°
(ii)

sinθ = 0.3827, then θ = 20° 30′
(iii)

sin θ = 0.6525, then θ = 40° 42’ + 2′ = 40°44′

#### Question 5.

Use tables to find the acute angle θ, if the value of cos θ is :
(i) 0.9848
(ii) 0.9574
(iii) 0.6885

From the tables of cosines, we find that if :
(i)

cos θ = 0.9848, then θ = 10°
(ii)

cos θ = 0.9574, then θ = 16°48′- 1’=16°47’
(iii)

cos θ = 0.6885, then θ = 46° 30′ or 46°30′
= 46° 29’

#### Question 6.

Use tables to find the acute angle θ, if the value of tan θ is :
(i) 2419
(ii) 0.4741
(iii) 0.7391
(iv) 1.06

From the table of tangents, we find that if:
(i)

tan θ = 0.2419, then θ=13° 36’
(ii)

tan θ = 0.4741, then θ = 25° 18’ + 4’ = 25°22′
(iii)

tan θ = 0.7391, then θ= 36°24’+ 4′ = 36°28′
(iv)

tan θ = 1.06, then θ = 46°36′ + 4′ = 46°40′

#### Question 7.

If sin θ=0.857; find:
(i) θ
(ii) tan θ

From the tables of T. Ratio’s we find this :
(i)

If sin θ = 0.857, then θ = 58°54′ + 4.5′ = 58° 58′ or 58°59’
(ii)

tan 58°58’= 1.6577 +43 = 1.662 or tan 58° 59′ = 1.6577 + 53 = 1.663

#### Question 8.

If θ is the acute angle and cos θ = 0.7258; find:
(i) θ
(ii) 2 tan θ – sin θ

From the tables of T-ratio’s, we find that:
(i)

If cos θ = 0.7258, then θ= 43° 30′ -2′ = 43°28’
(ii)

Now 2 tan θ – sin θ= 2 tan 43°28′ – sin 43°28′
2 tan 43°28’ = 2 x (0.9457 + 0.0022)
= 0:9479 x 2 = 1.8958
and sin 43°28′ = 0.6871 + 0.0008 = 0.6879
∴ 2 tan 43°28′ – sin 43° 28′ = 1.8958 – 0.6879 = 1.2079

#### Question 9.

Let θ be an acute angle and tan θ = 0.9490 find:
(i) θ
(ii) cos θ
(iii) sin θ – cos θ

From the tables of T-raios, we find that:
(i)

if tan θ = 0.9490 , then θ = 43°30′
(ii)

cos θ = cos 43°30′ = 0.7254
(iii)

sin θ = sin 43°50′ = 0.6884
∴ sin θ – cos θ = 0.6884 – 0.7254 = -0.0370 = -0.037

### Concise Selina Maths Solutions EXERCISE  – 21 (E)Trigonometrical Identities

#### Question 1.

Prove the following identities :

(i)…

(ii)…….

(iii)………

(iv)………

(v)………

(vi)……..

(vii)…….

(viii)…….

(ix)………

(x)……….

(xi)……….

(xii)……….

(xiii)……….

(xiv)……….

(xv)……….

(xvi)……….

(xvii)……….

#### Question 2.

If sin A + cos A = p and sec A + cosec A = q then prove that: q(p² – 1) 2p

………….

…………

#### Question 5.

If tan A=n tan B and sin A=m sin B, prove that:

………

#### Question 6.

(i) If 2 sin A-1 = 0, show that:
sin 3 A = 3 sin A – 4 sin3 A.             [2001]
(ii) If 4cos2 A-3 = 0, show that:
cos 3A = 4 cos3 A – 3 cos A

Evaluate:

(i)……

(ii)…….

(iii)………

(iv)………

(v)………

(vi)……..

(vii)…..

Prove that:

(i)……

(ii)…….

(iii)………

(iv)………

(v)………

#### Question 9.

If A and B are complementary angles, prove that:
(i) cot B + cos B sec A cos B (1 + sin B)
(ii) cot A cot B – sin A cos B – cos sin B = 0
(iii) cosec2 A + cosec2 B = cosec2 A cosec2 B

(iv) ………..

Prove that:

(i)…

(ii)…….

(iii)………

(iv)………

(v)………

(vi)……..

(vii)…….

(viii)…….

(ix)………

(x)……….

(x)

#### Question 11.

If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that :
(i) sin 3A = 3 sinA – 4 sin3A
(ii) cos 3A = 4 cos3 A – 3 cos A

#### Question 12.

Find A, if 0° ≤ A ≤ 90° and :
(i) 2 cos2 A – 1 = 0
(ii) sin 3A – 1 = 0
(iii) 4 sin2 A – 3 = 0
(iv) cos2 A – cos A = 0
(v) 2cos2 A + cos A – 1 = 0

#### Question 13.

If 0° < A < 90° ; find A, if :

………

#### Question 14.

Prove that : (cosec A – sin A) (sec A – cos A) sec2 A = tan A. (2011)

#### Question 15.

Prove the identity : (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ. (2014)

#### Question 16

Evaluate without using trigonometric tables,

sin2 28° + sin2 62° + tan2 38° – cot2 52° + sec2 30°

sin2 28° + sin2 62° + tan2 38° – cot2 52° + sec2 30°

= sin2 28° + [sin (90 – 28)°]2 + tan2 38° – [cot(90 – 38)°]2 + sec2 30°

= sin2 28°  + cos2 28° + tan2 38° – tan2 38° + sec2 30°

Thanks