ML Aggarwal ICSE Solutions Class 10 Math 18th Chapter Trigonometric Identities
Class 10 Chapter 18 Trigonometric IdentitiesChapter 18 – Trigonometric Identities
(1) If A is an acute angle and sin A = 3/5, find all other trigonometric ratios of angle A.
Solution : Given that,
Sin A = 3/5
In △ABC, ∠B = 90°
AC = 5cm and BC = 3cm
Then, by Pythagoras Theorem,
AC2 = AB2 + BC2
AB = AC2 – BC2
=> AB = √AC2 – BC2 = √25 – 9 = √16 = 4cm
Then,
{cos A = AB/AC = 4/5 tan A = BC/AB = ¾ sec A = 1/cosA = 5/4 cosec A = 1/sinA = 5/3}
(2) If A is an acute angle and sec A = 17/8, find all other trigonometric ratios of angle A.
Solution : Given that,
A is an acute angle secA = 17/8
In △ABC, AC = 17cm, AB = 8cm
By Pythagoras Theorem,
AC2 = AB2 + BC2
BC2 = AC2 – AB2
=> BC = √AC2 – AB2 = √289 – 64
BC = √225 = 15cm
Then,
{sinA = BC/AC = 15/17
cosA = 1/secA = 8/17
tanA = BC/AB = 15/8
cotA = 1/tanA = 8/15
cosecA = 1/sinA = 17/15}
(3) If 12cosecθ = 13, find the value of (2sinθ – 3cosθ)/(4sinθ – 9 cosθ)
Solution : Given that,
12 cosecθ = 13
cosecθ = 13/12
In △ABC = ∠A = θ
Then, cosecθ = AC/BC = 13/12
=> AC = 13 and BC = 12
By paythagoras Theorem,
AB = √AC2 – BC2 = √132 – 122 = √169 – 144 = √25 = 5
AB = 5cm
Then, sinθ = 1/cosecθ =12/13 and cosθ = AB/AC = 5/13
L.H.S. = (2 sinθ – 3cosθ)/(4sinθ – 9cosθ)
= 2 × 12/13 – 3 × 5/13 / 4 × 12/13 – 9 × 5/13
= (24 – 15)/48 – 45)
= 9/3
= 3
Thus, (2sinθ – 3cosθ)/(4sinθ – 9cosθ) = 3
(4) (i) cos2 (26°) + cos(64°), sin(26°) + (tan 36°/cot54°)
(ii) (sec17°/cosec73°) + (tan 68°/cot22°) + cos2 44° + cos246°)
Solution:
(i) Given that,
= cos2 (26°) + cos(64°). sin(26°) + (tan36°/cot54°)
= cos2 (26°) + cos(90 – 16). sin26° + [tan 36°/cot (90° – 54°)]
= [cos2(26°) + sin2 (26°)] + [tan36°/tan36°]
= 1 + 1 = 2
(ii) (sec17°/cosec 73°) + (tan 68°/cot22°) + cos2 44° + cos2 46°
= [(sec17°/cosec (90° – 73°)] + [tan (90° – 22°)/cot22°] + cos2 (90° – 44°)
= [sec17°/sec17°] + [cot22°/cot22°] + [sin246° + cos246° + cos246°]
= 1 + 1 + 1
= 3
(5) (i) (sin65°/cos25°) + (cos32°/sin88°) – sin28° – sec62° + cosec2 30°
Solution:
Given that,
(sin65°/cos25°) + (cos32°/sin88°) – sin28°. Sec62° + cosec230°
= [sin65°/cos(90°-65°)] + [cos32°/sin(90° – 32°)] – sin28°. Sec (90° – 28°)
= [sin65°/sin65°] + [cos32°/cos32°] – (sin28°× cosec28°] + 4
= 1 + 1 – 1 + 4
= 5
(ii) (sin29°/cosec61°) + 2cot8°. Cot17°. Cot45°. Cot73°. Cot82° – 3(sin238° + sin252°)
Solution:
Given that,
(sin29°/cosec61°) + 2cot8°.cot17°.cot45°.cot73°.cot82° – 3 (sin238° + sin252°)
= sin29°/cosec(90° – 29°) + [2 cot8°. Cot17°. Cot45°. Cot(90° – 17°). cot(90° – 8°) – 3 (sin238° + sin2(90° – 38°)]
= (sin29°/sin29°) + [2cot8°.cot17°. cot45°. tan17°.tan8°] – 3 [sin238° + cos238°]
= 1 + 2 [cot8° . tan8°) (cot17°. tan 17°) cot45°] – 3
= 1 + 2(1 × 1 × 1) – 3
= 1 + 2 – 3
= 0
(6.) (i.) (sin35°. Cos55° + cos35°. Sin55°) / (cosec210° – tan280°)
Solution:
Given that,
[(sin35°.cos55° + cos35°. Sin55°)] / (cosec2 10° – tan280°)
= [(sin35°. Cos(90° – 35°) + cos35°. sin(90° – 35°)] / (cosec210° – tan2(90° – 10°)]
= [sin35°. Sin35° + cos35°. Cos35°] / [cosec210° – cot210°]
= (sin235° + cos235°)/(cosec210° – cot210°) = (sin235° + cos235°)/(cosec210° – cot210°) = 1/1
= 1
(ii) sin234° + sin256° + 2tan18°. Tan72° -cot230°
Solution: Given that,
sin234° + sin256° + 2tan18°. Tan72° – cot230°
= sin234° + sin2(90° – 34°) + 2tan18°. Tan (90° – 18°) – cot230°
= (sin234° + cos234°) + 2tan18°.cot18° – cot230°
= 1 + 2 × 1 – (√3)2
= 1 + 2 – 3
= 0
(8) Prove that:
(i) (secA + tanA) (1 – sinA) = cosA
(ii) (1 + tan2A) (1 – sinA) (1 + sinA) = 1
Solution:
(i) L.H.S. = (secA + tanA) (1 – sinA)
= (1/cosA + sinA/cosA) (1 – sinA)
= (1 + sinA/cosA) (1 – sinA)
= (1 + sinA) (1 – sinA)/cosA = (1 – sin2A)/cosA
= cos2A/cosA ∵ sin2A + cos2A = 1
= cosA
L.H.S. = R.H.S.
(ii) L.H.S. = (1 + tan2A) (1 – sinA) (1 + sinA)
= (1 + sin2A/cos2A) (1 – sin2A) ∵ (a+b) (a-b) = a2 – b2
= (cos2A + sin2A)/cos2A (1 – sin2A)
= 1/cos2A (cos2A)
= 1 ∵ sin2A + cos2A = 1
L.H.S. = R.H.S.
(9) (i) tanA + cotA = secA . cosecA
(ii) (1 – cosA) (1 + secA) = tanA. sinA
Solution: (i) L.H.S. = tanA + cotA
= sinA/cosA + cosA/sinA ∵ 1/sinA = cosecA
= (sin2A + cos2A)/sinA . cosA = 1/sinA. cosA ∵ 1/cosA = secA
= 1/sinA . 1/cosA = cosecA . secA ∵ sin2A + cos2A = 1
= secA . cosecA
L.H.S. = R.H.S.
(ii) (1 – cosA) (1 + secA)
= (1 – cosA) (1 + 1/cosA) ∵ secA = 1/cosA
= (1 – cosA) (cosA + 1)/cosA
= (1 – cosA) (1 + cosA)/cosA
= (1 – cos2A) . 1/cosA ∵(a-b) (a+b) = a2 – b2
= sin2A. 1/cosA ∵ sin2A + cos2A = 1
= sinA/cosA . sinA
(1 – cosA) (1 + secA) = tanA . sinA
(10) (i) cot2A – cos2A = cot2A . cos2A
(ii) 1 + tan2A/(1 + secA) = secA
(iii) (1 + secA)/secA = sin2A/(1 – cosA)
(iv) sinA/(1 – cosA) = cosecA + cotA
Solution: (i) cot2A – cos2A = cos2A/sin2A – cos2A ∵ cotA = cosA/sinA
= (1/sin2A – 1)cos2A
= (1 – sin2A)/sin2A – cos2A
= cos2A/sin2A – cos2A ∵ sin2A + cos2A = 1
cot2A – cos2A = cot2A . cos2A Hence proved.
(ii) 1 + tan2A/(1 + secA) = secA
1 + tan2A/1 + secA = 1 + (sec2A – 1)/(1 + secA) ∵ 1 + tan2A =sec2A tan2A = sec2A – 1
= 1 + secA + sec2A – 1/1 + secA
= secA(1 + secA)/(1 + secA)
1 + tan2A/(1 + secA) = secA Hence proved.
(iii) (1 + secA)/secA = 1/secA + 1
= cosA + 1
= (1 + cosA) (1 – cosA)/(1 – cosA)
(1 + secA)/secA = sin2A/(1 – cosA) Hence proved.
(iv) sinA/(1 – cosA) = sinA (1 + cosA)/(1 – cosA) (1 + cosA) multiply Dor& Nor by (1 + cosA)
= sinA (1 + cosA)/(1 – cos2A)
= sinA(1 + cosA)/sin2A ∵ cos2A + sin2A = 1
= 1/sinA(1 + cosA)
= 1/sinA + cosA/sinA
sinA/(1 – cosA) = cosecA + cotA Hence proved.
(11.) (i) sinA/(1 + cosA) = (1 – cosA)/sinA
Solution:
sinA/(1 + cosA) = sinA (1 – cosA)/(1 + cosA) (1 – cosA) ∵ multiply Dor& Nor by (1 – cosA)
= sinA (1 – cosA)/(1 – cos2A) ∵ (a – b) (a + b) = a2 – b2
= sinA (1 – cosA)/sin2A ∵ sin2A + cos2A = 1
sinA/(1 + cosA) = (1 – cosA)/sinA Hence proved.
(ii.) (1 – tan2A)/(cot2A – 1) = tan2A
(1 – tan2A)/(cot2A – 1) = 1 – sin2A/cos2A / cos2A/sin2A – 1
= (cos2A – sin2A)/cos2A / (cos2A – sin2A)/sin2A
= 1/cos2A / 1/sin2A = sin2A/cos2A = tan2A Hence proved.
(iii.) sinA/(1 + cosA) = cosecA – cotA
Solution: sinA/(1 + cosA) = sinA(1 – cosA)/(1 + cosA) (1 – cosA)
= sinA (1 – cosA)/(1 – cos2A)
= sinA (1 – cosA)/sin2A ∵ sin2A + cos2A = 1
= (1 – cosA)/sinA = 1/sinA – cosA/sinA
sinA/(1 + cosA) = cosecA – cotA Hence proved.
(12) (i) (secA – 1)/(secA + 1) = (1 – cosA)/(1 + cosA)
Solution: (secA – 1)/(secA + 1) = 1/cosA – 1 / 1/cosA + 1
= (1 – cosA)/cosA / (1 + cosA)/cosA
(secA – 1)/(secA + 1) = (1 – cosA)/(1 + cosA) Hence proved.
(ii) sec2A + cosec2A = sec2A. cosec2A
Solution: sec2A + cosec2A = 1/cos2A + 1/sin2A
= (sin2A + cos2A)/sin2A . cos2A
= 1/sin2A . cos2A
= 1/sin2A . 1/cos2A
= cosec2A. sec2A
(sec2A + cosec2A) = sec2A . cose2A Hence proved.
(13) (i) (1 + sinA)/cosA + cosA/(1 + sinA) = 2secA
Solution:
(1 + sinA)/cosA + cosA/(1 + sinA) = (1 + sinA)2 + cos2A/cosA (1 + sinA)
= 1 + 2sinA + sin2A + cos2A/ cosA + cosA.sinA
= 1 + 2sinA + 1/ cosA(1 + sinA)
= 2 + 2sinA/ cosA(1 + sinA)
= 2 (1 + sinA)/ cosA(1 + sinA)
= 2/cosA
(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA Hence proved.
(ii) tanA/(secA – 1) + tanA/(secA + 1) = 2cosecA
Solution:
tanA/(secA – 1) + tanA/(secA + 1) = tanA [2/secA – 1 + 1/secA + 1]
= tanA [secA + 1 + secA – 1/(secA + 1) (secA – 1)]
= tanA [2secA/(sec2A – 1)]
= 2tanA . secA/tan2A ∵ 1 + tan2A = sec2A
= 2secA/tanA
= 2/cosA / sinA/cosA
= 2/sinA
tanA/(secA – 1) + tanA/(secA + 1) = 2 cosecA Hence proved.
(14) (i) cotA – tanA = (2cos2A – 1) / (sinA + cosA)
Solution: cotA – tanA = cosA/sinA – sinA/cosA
= cos2A – sin2A/sinA. cosA
= cos2A – (1 – cos2A)/sinA.cosA ∵ sin2A + cos2A = 1
= cos2A – 1 + cos2A/sinA.cosA
cotA – tanA = (2 cos2A – 1)/sinA.cosA Hence proved.
(ii) cotA – 1 / 2 – sec2A = cotA/1 + tanA
Solution: cotA – 1 / 2 – sec2A = cosA/sinA – 1 / 2 – 1/cos2A
= (cosA – sinA)/sinA / (2cos2A – 1)/cos2A = cos2A(cosA – sinA)/sinA (2cos2A – 1)
= cos2A (cosA – sinA)/sinA [2cos2A – sin2A – cos2A]
= cos2A(cosA – sinA)/sinA (cos2A – sin2A) = cos2A(cosA – sinA)/sinA (cosA – sinA) (cosA + sinA)
= cos2A/sinA (cosA + sinA)
= cosA/sinA / (cosA + sinA)/cosA
= cotA/(1 + sinA/cosA)
cotA – 1 / 2 – sec2A = cotA/1 + tanA Hence proved.
(15) (i) tan2θ – sin2θ = tan2θ.sin2θ
Solution: tan2θ – sin2θ = sin2θ/cos2θ – sin2θ
= sin2θ – sin2θ(cos2θ)/cos2θ
= sin2θ (1 – cos2θ)/cos2θ
= sin2θ/cos2θ (1 – cos2θ) ∵ sin2θ + cos2θ = 1
tan2θ – sin2θ = tan2θ.sin2θ Hence proved.
(ii) cosθ/(1 – tanθ) – sin2θ/(cosθ – sinθ) = cosθ + sinθ
Solution: cosθ/(1 – tanθ) – sin2θ/(cosθ – sinθ) = cosθ/(1 – sinθ/cosθ) – sin2θ/(cosθ – sinθ)
= cosθ/(cosθ – sinθ)/cosθ – sin2θ/(cosθ – sinθ)
= cos2θ/cosθ – sinθ) – sin2θ/(cosθ – sinθ)
= (cos2θ – sin2θ)/(cosθ – sinθ)
= (cosθ + sinθ) (cosθ – sinθ)/cosθ – sinθ) ∵ (a-b) (a+b) = a2 – b2
cosθ/(1 – tanθ) – sin2θ/cosθ – sinθ) = cosθ + sinθ Hence proved.
(16) (i) cosec4θ – cosec2θ = cot4θ + cot2θ
Solution: cosec4θ – cosec2θ = cosec2θ (cosec2θ – 1)
= cosec2θ.cot2θ 1 + cot2θ = cosec2θ
= (1 + cot2θ). Cot2θ
= cot2θ + cot4θ
cosec4θ – cosec2θ = cot4θ + cot2θ Hence proved.
(ii) 2sec2θ – sec4θ – 2cosec2θ + cosec4θ = cot4θ – tan4θ
Solution:
= 2sec2θ – sec4θ – 2cosec2θ + cose4θ
= 2 (1 + tan2θ) – (1 + tan2θ)2 – 2 (1 + cot2θ) + (1 + cot2θ)2
= 2 tan2θ + 2 – (tan4θ + 2tan2θ + 1) – 2cot2θ – 2 + (cot4θ + 2cot2θ + 1)
= cot4θ – tan4θ
= R.H.S.
(17) (i) (1 + cosθ – sin2θ)/sinθ (1 + cosθ) = cotθ
Solution: 1 + cosθ – sin2θ/sinθ(1 + cosθ) = (1 – sin2θ) + cosθ/sinθ (1 + cosθ)
= cos2θ + cosθ/(1 + cosθ) sinθ∵ sin2θ + cos2θ = 1
= cosθ/sinθ (1 + cosθ)/(1 + cosθ)
1 + cosθ – sin2θ/sinθ (1 + cosθ) = cotθ Hence proved.
(ii) (tan3θ – 1)/(tanθ – 1) = sec2θ + tanθ
Solution: tan3θ – 1/tanθ – 1 = (tan3θ – 13)/tanθ – 1) ∵ (a3 – b3) = (a – b) (a2 + ab + b2)
= (tanθ – 1)/(tanθ – 1) (tan2θ + tanθ + 1)
= (1 + tan2θ + tanθ) ∵ 1 + tan2θ = sec2θ
tan3θ – 1/tanθ – 1 = (sec2θ + tanθ) Hence proved.
(18) (i) (1 + cosecA)/cosecA = cos2A/(1 – sinA)
Solution: 1 + cosecA/cosecA = 1 + 1/sinA / 1/sinA
= (sinA + 1)/sinA / 1/sinA
= (1 + sinA) (1 – sinA)/(1 – sinA)
= (1 – sin2A)/(1 – sinA) = cos2A/(1 – sinA) ∵ cos2A + sin2A = 1
(1 + cosecA)/cosecA = cos2A/(1 – sinA)
Hence proved.
(ii) √(1 – cosA)/(1 + cosA) = cosecA – cotA
Solution: √(1 – cosA)/(1 + cosA) = √(1 – cosA) (1 – cosA)/(1 + cosA) (1 – cosA)
= √(1 – cosA)2/(1 – cos2A)
= √(1 – cosA)2/sin2A
= 1 – cosA/sinA
= 1/sinA = cosA/sinA
= cosecA – cotA Hence proved.
(19) (i) √1 + sinA/1 – sinA = tanA + secA
Solution:√(1 + sinA)/(1 – sinA) = √(1 + sinA) (1 + sinA)/(1 – sinA) (1 + sinA)
= √(1 + sinA)2/(1 – sin2A)
= √(1 + sinA)2/cos2A = √(1 + sinA)2/cos2A = (1 + sinA)/cosA
= 1/cosA + sinA/cosA
= secA + tanA
√(1 + sinA/(1 – sinA) = tanA + secA Hence proved.
(ii) √(1 – cosA)/(1 + cosA) = cosecA – cotA
Solution :√(1 – cosA)/(1 + cosA) = √(1-cosA) (1 – cosA)/(1 + cosA) (1 – cosA)
= √(1 – cosA)2/(1 – cos2A)
= √(1 – cosA)2/sin2A = (1 – cosA)/sinA
= 1/sinA – cosA/sinA
√(1-cosA)/(1+cosA) = cosecA – cotA Hence Proved
(20) (i) √secA – 1/secA + 1 + √secA + 1/secA – 1 = 2 cosecA
Solution:
√secA – 1/secA + 1 + √secA + 1/secA – 1 = √(secA – 1) (secA – 1)/(secA + 1) (secA – 1) + √(secA + 1) (secA + 1)/(secA – 1) (secA + 1)
= √(secA – 1)2/(sec2A – 1) + √(secA + 1)2/(sec2A – 1)
= √(secA – 1)2/tan2A + √(secA + 1)2/tan2A ∵ 1 + tan2A = sec2A
= (secA – 1)/tanA + (secA + 1)/tanA = secA – 1 + secA + 1/tanA
= 2secA/tanA = 2/cosA / sinA/cosA
= 2/sinA = 2cosecA
L.H.S. = R.H.S. Hence proved.
(ii) cosA.cotA/(1 – sinA) = 1 + cosecA
Solution:
cosA.cotA/(1 – sinA) = cosA . cosA/sinA / (1 – sinA)
= cos2A/sinA / (1 – sinA) = cos2A/sinA(1 – sinA)
= 1 – sin2A/sinA (1 – sinA) = (1 – sinA) (1 + sinA)/sinA (1 – sinA)
= (1 + sinA)/sinA
= 1/sinA + 1
= cosecA + 1
cosA.cotA/(1 – sinA) = cosecA + 1 Hence proved.
(23) cotθ + cosecθ – 1/(cotθ – cosecθ + 1) = (1 + cosθ)/sinθ
Solution:
(cotθ – cosecθ – 1)/(cotθ – cosecθ + 1) = cosθ/sinθ + 1/sinθ – 1 / cosθ/sinθ – 1/sinθ + 1
= cosθ + 1 – sinθ/cosθ – 1 + sinθ
= cosθ + (1 – sinθ)/cosθ – (1 – sinθ)
= [cosθ + (1 – sinθ)]/[cosθ – (1 – sinθ)] [cosθ + (1 – sinθ)]/[cosθ + (1 – sinθ)]
= [cosθ + (1 – sinθ)]2/cos2θ – (1 – sinθ)2
= cos2θ + (1 – sinθ)2 + 2cosθ (1 – sinθ)/cos2θ – (1 – 2sinθ + sin2θ)
= cos2θ + 1 – 2sinθ + sin2θ + 2cosθ – 2sinθ . cosθ/cos2θ – 1 + 2sinθ – sin2θ
= 2 – 2sinθ + 2cosθ – 2sinθ.cosθ/1 – sin2θ – 1 + 2sinθ – sin2θ
= 2 + 2cosθ – 2sinθ – 2sinθ . cosθ/2sinθ – 2sin2θ
= 2(1 + cosθ) – 2sinθ(1 + cosθ)/2sinθ (1 – sinθ)
= (1 + cosθ) (2 – 2sinθ)/2sinθ (1 – sinθ) = 2(1 + cosθ) (1 – sinθ)/2sinθ (1 – sinθ)
= 1 + cosθ/sinθ
L.H.S. = R.H.S. Hence proved.
(24) (sinθ + cosθ) (secθ + cosecθ) = 2 + secθ. cosecθ
Solution:
(sinθ + cosθ) (secθ + cosecθ) = (sinθ + cosθ) (1/cosθ + 1/sinθ)
= (sinθ + cosθ) (sinθ + cosθ)/sinθ . cosθ
= (sinθ + cosθ)2/sinθ . cosθ
= sin2θ + cos2θ + 2sinθ . cosθ/sinθ . cosθ
= 1 + 2sinθ.cosθ/sinθ . cosθ
= 1/sinθ . 1/cosθ + 2
= Cosecθ.secθ + 2
(sinθ + cosθ) (secθ + cosecθ) = 2 + secθ . cosecθ Hence proved.
(25) tan2A/1 + tan2A + cot2A/1 + cot2A = 1
Solution:
tan2A/(1 + tan2A) + cot2A/(1 + cot2A) = tan2A/(1 + tan2A) + 1/tan2A / 1 + 1/tan2A
= tan2A/(1 + tan2A) + 1/tan2A / (tan2A + 1)/tan2A
= (tan2A)/(1 + tan2A) + 1/(1 + tan2A)
= (1 + tan2A)/(1 + tan2A)
= 1
L.H.S. = R.H.S. Hence proved.
(26) (i) 1/secA + tanA – 1/cosA = 1/cosA – 1/(secA – tanA)
Solution:
1/(secA + tanA) – 1/cosA = 1/ 1/cosA + sinA/cosA – 1/cosA
= cosA/(1 + sinA) – 1/cosA
= cos2A – (1 + sinA)/cosA (1 + sinA)
= 1/(1/cosA + sinA/cosA) – 1/cosA
= 1/cosA + 1/(secA + tanA)
L.H.S. = R.H.S. Hence proved.
(ii) (tanA + sinA)/(tanA – sinA) – (secA + 1)/(secA – 1)
Solution:
(tanA + sinA)/(tanA – sinA) = (sinA/cosA + sinA) / (sinA/cosA – sinA)
= sinA (1/cosA + 1)/sinA (1/cosA – 1)
= (secA + 1)/(secA – 1)
L.H.S. = R.H.S. Hence proved.
(27) If (sinθ + cosθ) = √2sin (90 – θ), show that cotθ = √2 + 1
Solution:
Given that,
sinθ + cosθ = √2 sin(90 – θ)
sinθ + cosθ = √2 cosθ
=> (sinθ + cosθ)/sinθ = √2 cosθ/sinθ
1 + cotθ = √2 cotθ
1 = (√2 – 1) cotθ
cotθ = 1/(√2 – 1) = (√2 + 1)/(√2 -1) (√2 + 1) = √2 + 1/(2 – 1)
cotθ = √2 + 1 Hence proved.
(28) If 7(sin2θ) + 3cos2θ = 4, θ°≤ θ ≤ 90°, then find the value of θ.
Solution:
Given that,
7sin2θ + 3cos2θ = 4, θ≤θ≤ 90
3sin2θ + 4sin2θ + 3cos2θ = 4
(3sin2θ + 3cos2θ) + 4sin2θ = 4
3(sin2θ + cos2θ) + 4sin2θ = 4
3 + 4sin2θ = 4
4sin2θ = 1
sin2θ = ¼
sinθ = ½
θ = sin-1(1/2)
θ = 30° is the required value of θ.
(29) If (secθ + tanθ) = m and (secθ – tanθ) = n, prove that m × n = 1.
Solution:
Given that,
(secθ + tanθ) = m and (secθ – tanθ) = n
L.H.S. = m × n
= (secθ + tanθ) (secθ – tanθ)
= sec2θ – tan2θ∵ 1 + tan2θ = sec2θ
= 1
= R.H.S.
mn = 1 Hence proved.
(30) If x = asecθ + btanθ and y = atanθ + bsecθ, prove that x2 – y2 = a2 – b2
Solution:
Given that,
x = asecθ + btanθ & y = atanθ + bsecθ
Then, L.H.S. = x2 – y2
= (asecθ + btanθ)2 – (atanθ + bsecθ)2
= a2sec2θ + b2tan2θ + 2absecθ.tanθ – (a2tan2θ + b2sec2θ + 2abtanθ.secθ)
= a2sec2θ + b2tan2θ + 2absecθ.tanθ – a2tan2θ – b2sec2θ – 2absecθ tanθ
= a2(sec2θ – tan2θ) + b2 (tan2θ – sec2θ)
= a2 (1) + b2(-1) ∵ 1 + tan2θ = sec2θ
= a2 – b2
= R.H.S.
Thus, x2 – y2 = a2 – b2 Hence proved.