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.
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EXERCISE -21(A) Trigonometrical Identities Concise Selina Maths Solutions for ICSE Class 10
Prove the following Identities :
Question 1.
…………
Answer 1
Question 2.
Answer 2
Question 3.
……….
Answer 3
Question 4.
………..
Answer 4
Question 5.
sin4A – cos4 A = 2 sin2A-1
Answer 5
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
Answer 6
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
Answer 7
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
Answer 8
Question 9.
cosec A (1 + cos A) (cosec A – cot A) = 1
Answer 9
Question 10.
sec2 A + cosec2A = sec2 A cosec2 A
Answer 10
Question 11.
…………
Answer 11
Question 12.
tan2A – sin2A = tan2 A. sin2 A
Answer 12
Question 13.
cot2 A – cos2 A = cos2 A. cot2 A
Answer 13
Question 14.
(cosecA + sinA) (cosec A – sinA) = cot2 A + cos2A
Answer 14
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
Answer 15
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
Answer 16
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
Answer 17
Question 18.
……
Answer 18
Question 19.
……….
Answer 19
Question 20.
………..
Answer 20
Question 21.
…………..
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Answer 21
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 + cos2 A+sec2A + 2cosA x
= 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
Answer 22
Question 23.
……..
Answer 23
Question 24.
………..
Answer 24
Question 25.
…………
Answer 25
Question 26.
………….
Answer 26
Question 27.
………….
Answer 27
Question 28.
…………
Answer 28
Question 29.
…………..
Answer 29
Question 30.
…………..
Answer 30
Question 31.
……………
Answer 31
Question 32.
………….
Answer 32
Question 33.
……………
Answer 33
Question 34.
……………
Answer 34
Question 35.
………………
Answer 35
Question 36.
…………..
Answer 36
Question 37.
………….
Answer 37
Question 38.
(1 +cot A-cosec A) (1 + tan A + sec A) = 2
Answer 38
Question 39.
………..
Answer 39
Question 40.
……………
Answer 40
Question 41.
…………..
Answer 41
Question 42.
………….
Answer 42
Question 43.
………….
Answer 43
Question 44.
…………
Answer 44
Question 45.
…………
Answer 45
Question 46.
…………..
Answer 46
Question 47.
……………
Answer 47
Question 48.
…………..
Answer 48
EXERCISE – 21 (B) Solutions of Selina Concise Mathematics Chapter-20 Trigonometrical Identities
Question 1.
Prove that:
(i)…
(ii)…….
(iii)………
(iv)………
(v)………
(vi)……..
(vii)…….
(viii)…….
(ix)……….
Answer 1
(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
Answer 2
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
Adding we get,
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
Answer 3
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-b2 ( ∵ 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.
Answer 4
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
Adding, we get,
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
Answer 5
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
Answer 6
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
Adding, we get
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.
Question 7.
…………
Answer 7
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) ………
Answer 1
(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°
Answer 2
(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°
Question 3.
Show that:
…………
Answer 3
Question 4.
For triangle ABC, Show that:
Answer 4
Question 5.
Evaluate:
(i)…
(ii)…….
(iii)………
(iv)………
(v)………
(vi)……..
(vii)…….
(viii)…….
(ix)……….
Answer 5
(i)
Question 6.
………..
Answer 6
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°
Answer 7
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
Answer 8
Question 9.
Prove that:
…………….
Answer 9
Question 10.
Evaluate:
………….
Answer 10
Question 11.
Without using trigonometric tables, evaluate sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°.
Answer 11
Question 12.
Without using trigonometrical tables, evaluate: cosec2 57° – tan2 33° + cos 44° cosec 46° – cos45°- tan2 60°
Answer 12
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′
Answer 1
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′
Answer 2
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′
Answer 3
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
Answer 4
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
Answer 5
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
Answer 6
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
Answer 7
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
Answer 8
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
Answer 9
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)……….
Answer :1
Answer (i)
Answer :(ii)
Answer :(iii)
Answer :(iv)
Answer :(v)
Answer :(vii)
Answer :(viii)
Answer :(ix)
Answer :(x)
Answer :(xi)
Answer :(xii)
Answer :(xiii)
Answer :(xiv)
Answer :(xv)
Answer :(xvi)
Answer (xvii)
Question 2.
If sin A + cos A = p and sec A + cosec A = q then prove that: q(p² – 1) 2p
Answer 2
Question 3.
………….
Answer 3
Question 4.
…………
Answer 4
Question 5.
If tan A=n tan B and sin A=m sin B, prove that:
………
Answer 5
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
Answer 6
Question 7.
Evaluate:
(i)……
(ii)…….
(iii)………
(iv)………
(v)………
(vi)……..
(vii)…..
Answer 7
Question 8.
Prove that:
(i)……
(ii)…….
(iii)………
(iv)………
(v)………
Answer 8
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) ………..
Answer 9
Question 10.
Prove that:
(i)…
(ii)…….
(iii)………
(iv)………
(v)………
(vi)……..
(vii)…….
(viii)…….
(ix)………
(x)……….
Answer 10
(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
Answer 11
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
Answer 12
Question 13.
If 0° < A < 90° ; find A, if :
………
Answer 13
Question 14.
Prove that : (cosec A – sin A) (sec A – cos A) sec2 A = tan A. (2011)
Answer 14
Question 15.
Prove the identity : (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ. (2014)
Answer 15
Question 16
Evaluate without using trigonometric tables,
sin2 28° + sin2 62° + tan2 38° – cot2 52° + sec2 30°
Answer 16
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°
End of Trigonometrical Identities Concise Solutions Chapter-21
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