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Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes Revision Question Bank Introduction to Trigonometry 1. Evaluate sec2 900 cot 2 2 sin2 250 sin2 650 2cos2 600 tan2 28 tan2 620 cot 400 tan 500 3 sec2 430 cot 2 470 Solution : sec2 90 cot cos2 60o tan2 28o tan2 62o cot 40o + tan50o 3 sec2 43o cot 2 47o 2 sin2 25o sin2 65o 1 2 tan2 28tan2 90o 28o cosec cot cot 40o 2 = + + o o 2 sin2 25o sin2 90o 43o 3 sec2 43o cot 2 90o 43o tan 90 40 2 2 sec 90o cosec 1 2 o 2 o 2 tan 28 cot 28 cosec cot cot 40o 4 = o 2 sin2 cos2 25o 3 sec2 43o tan2 43o cot 40 2 2 sin 90o cos ,tan 90o cot and cot 90o tan 1 1 1 2 = 2 1 3 3 1 2 2 2 2 2 2 1 cosec cot 1,sin cos 1,sec tan 1 and tan cot = 2. 1 1 3 1 6 10 5 1 2 6 6 6 3 Show that: 1 1 2sec2 . 1 sin 1 sin Solution : LHS. = 1 1 1 sin 1 sin 1 sin 1 sin 1 sin 1 sin www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 3 Pioneer Education {The Best Way To Success} = 2 1 sin2 = 2 2sec2 2 cos IIT – JEE /AIPMT/NTSE/Olympiads Classes a2 b2 a b a b sin2 cos2 1 LHS. = RHS. 3. Prove the identity (1 + cosec ) (1 – sin ) = cos cot . Solution : To prove, 1 cosec1 sin = cos cot LHS. = 1 cosec 1 sin 1 = 1 1 sin sin = sin 1 . 1 sin 1 sin sin sin cos2 = sin = 1 cosec sin sin2 cos2 1 cos .cos sin cos cot sin = cot .cos LHS=RHS 4. Hence proved Without using trigonometric tables, evaluate: cos2 200 cos2 700 × sec2 600 – 2 cot580cot 320 – 4 tan 130 tan 370 tan 450 tan 530 tan 770 sec2 500 cot 2 400 Solution : cos2 20o cos2 70o 2sec2 60o 2cot58o cot32o 4tan13o tan37o tan45o tan53o tan77o 2 o 2 o sec 50 cot 40 22 2cot58 cot 90 sec 50 cot 90 50 1 tan 90 37 tan 90 13 cos2 20o cos2 90o 20o 2 o 2 o o o 2 o o o o 58o 4tan13o tan37o o www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 4 Pioneer Education {The Best Way To Success} = IIT – JEE /AIPMT/NTSE/Olympiads Classes cos 20 sin 20 8 2cot58o tan58o 2 o 2 o sec 50 tan 50 2 o 2 o 4tan13o tan37o cot37o cot13o cos 90o sin ,ta 90o cot = 1 8 2 1 4 1 1 1 tan cot 1 =8–2–4=2 5. Prove the identity: 1 cos A 1 cos A 4 cot A cosec . 1 cos A 1 cos A Solution : To prove, 1 cos A 1 cos A 4cot AcosecA 1 cos A 1 cos A LHS = 1 cos A 1 cos A 1 cos A 1 cos A 1 cos A 1 cos A = 1 cos A 1 cos A 2 = 2 1 cos A 1 cos A 1 cos A 1 cos A 1 cos A 2 = 22cos A sin2 A = 4cos A 1 . 4cot A.cosesA sin A sinA a2 b2 a b a b sin2 A cos2 A 1 cos A 1 ,cosecA = cot A sin A sin A LHS =RHS 6. Hence proved Show that: (cosec A –1) (cosec A +1) (sec A –1) (sec A +1) =1. Solution : LHS www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 5 Pioneer Education {The Best Way To Success} = cosecA 1 cosecA 1 sec A 1 sec A 1 IIT – JEE /AIPMT/NTSE/Olympiads Classes 1 1 1 1 = 1 1 1 1 sin A sin A cos A cos A 1 cosecA ,sec A sin A cos A 1 sin A 1 sin A 1 cos A 1 cos A = sin A sin A cos A cos A 1 sin A . 1 cos A = 2 2 sin A = 2 a2 b2 a b a b 2 cos A cos2 A sin2 A .cos sin2 A sin2 A sin2 A cos2A 1 =1 LHS=RHS 7. Show that: 2 cosec2A sin A cos A . cosec2A 2cot A sin A cos A Solution : 1 2 cosec A sin2 A LHS = 1 2cos A cosec2A 2cot A 2 sin A sin A 2 2 1 cos A cosecA and cot A sin A sin A 2sin2 A 1 sin2 A = 1 2sin Acos A sin2 A 2sin2 A 1 sin2 A = Sin2A 1 2sin Acos A 2sin2 A 1 = sin2 A cos2 A 2sin Acos A sin2 A cos2 A 1 www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 6 Pioneer Education {The Best Way To Success} 2sin A sin A cos A 2 = = = 2 2 sin A cos A IIT – JEE /AIPMT/NTSE/Olympiads Classes 1 sin2 cos2 2sin2 A sin2 A cos2 A sin A cos A 2 sin2 A cos2 A sin A cos A 2 = sin A cos A sin A cos A 2 sin A cos A = sin A cos A sin A cos A a2 b2 a b a b LHS=RHS 8. If 3 cot A = 4 , then check whether 1 tan2 A cos2 A sin2 A or not. 2 1 tan A Solution : Let us consider a right angled ABC in which B 90o . For A , Base = AB and perpendicular = BC. Also, hypotenuse = AC 3cot A 4 cot A 4 3 But cot A = ……..(i) Base AB perpendicular BC From(i) and (ii) we get AB 4 4k BC 3 3k AB = 4k and BC = 3k Using Pythagoras theorem AC2 = AB2 + BC2 AC2 = (4k)2 + (3k)2 www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 7 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes AC = 16k 2 9k 2 = 25k 2 5k Now, sin A= BC 3k 3 AC 5k 5 cos A = 2 5k AB 4k 4 AC 5k 5 Also, tan A = BC 3k 3 AB 4k 4 Now, to check the given equation, 2 3 1 1 9 2 1 tan A 4 16 LHS 2 2 1 tan A 3 1 9 1 16 4 16 9 7 7 = 16 16 16 9 25 25 16 16 RHS = cos2 A sin2 A 2 2 4 3 16 9 16 9 7 = 25 25 5 5 25 25 From Eqs. (i) and (ii), LHS=RHS 1 tan2 A cos2 A sin2 A 2 1 tan A 9. Prove the identity (cot A – tan A) cos A = cosec A – 2 sin A. Solution : To prove (cot A – tan A) cos A = cosec A – 2sin A LHS = (cot A – tan A) cos A cos A sin A = cos A sin A cos A www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 8 Pioneer Education {The Best Way To Success} cos 2 = = A sin A 2 sin Acos A .cos A IIT – JEE /AIPMT/NTSE/Olympiads Classes cos2 A sin2 A sin A cos2 A sin A = sin A 1 sin2 A = sin A sin A = 1 sin A sin A cos2 1 sin2 sin A = cosec A – 2 sin A LHS = RHS Hence proved 1 1 10. Show that cos can be written in the form k tan and find the 1 sin 1 sin value of k. Solution : 1 1 cos 1 sin 1 sin 1 sin 1 sin = cos 1 sin 1 sin 2sin = cos 2 1 sin = cos . a2 b2 a b a b 2sin sin 2 2 cos cos = 2tan k 2 sin tan cos www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 9 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes Chapter Test {Trigonometry} M: Marks: 40 M: Time: 40 Min. 1. Without using trigonometrical tables, evaluate cos580 sin 220 cos 380 sin320 cos 680 tan180 tan350 tan600 tan720 tan550 [4] Solution: cos 580 sin 220 sin 320 cos 680 cos 380 tan 180 tan350 tan 600 tan 720 tan550 = cos 900 320 sin 900 680 sin 320 cos 680 cos 380 cosec 900 380 tan180 tan350 3 tan 900 380 tan 900 350 cos 380 sec 380 sin 320 cos 680 = – sin 320 cos 680 tan 180 tan 350 3 cot 180 cot350 cos 900 sin , sin 900 cos , cosec 900 sec 0 and tan 90 cot = 11 [ tan 18 cot18 0 0 1 tan 350 cot 350 3 cos . sec 1 ] = 11 = 1 1.1. 3 2 3 1 3 www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 10 Pioneer Education {The Best Way To Success} 2. If IIT – JEE /AIPMT/NTSE/Olympiads Classes sin cos 1 sin cos 1 3 , then prove that . 3 3 a b a b ab a b 4 8 4 8 [4] Solution: sin8 cos8 1 3 To prove 3 3 a b a b sin4 cos4 1 Given, a b ab 4 4 sin cos 2 a b 1 b a sin4 cos4 2 2 a b sin cos b a ab 4 ab 4 sin cos a b 2 [ sin2 cos2 1 ] = sin4 cos 4 sin 4 cos 2 b a sin4 cos4 cos4 a b = sin4 cos4 2sin 2 cos 2 b a sin4 cos4 2 sin2 cos2 0 a b Þ sin4 2 2 b a b a sin2 cos2 – 2 sin2 cos2 0 a a a a 2 b a sin2 cos2 0 b a b a sin2 cos2 0 a b sin cos2 2 [ a2 b2 2ab a b ] 2 [taking square root] a b a b b a sin2 cos2 k say a b sin2 ak and cos2 bk ..(i) www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 11 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes sin cos ak bk k ab ab From Eqs. (i) and (ii), we get sin2 cos2 sin2 cos2 a b ab sin2 cos2 1 a b ab Taking first and third terms, sin2 1 a sin2 a ab ab Taking second and third terms, cos2 1 b ab b cos2 ab 2 2 2 sin8 cos8 sin Now, 3 a3 b a3 4 a ab ab ab = 3 3 a b a b = 4 4 a b a b = ab a b 4 ..(ii) .(iii) ..(iv) cos 4 2 4 b3 4 [ from Eqs (iii) and (iv)] 1 a b 3 3. If 7 sin2 3cos2 4 and is an acute angle, then prove that: sec cosec 2 2 . [4] 3 Solution: Given, 7sin2 3cos2 4 4sin2 3sin2 3 cos2 4 4sin2 3 sin2 cos2 4 4sin2 3 4 [ sin2 cos2 1 ] 4sin2 4 3 4sin2 1 www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 12 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes 1 4 1 sin 2 [taking positive square root as is acute angle] sin2 cosec 2 sin 1 cosec cos 1 sin2 and 1 = 1 2 = 1 = 2 1 4 4 1 4 3 3 4 2 sec 2 3 sec cosec sec 1 cos 2 2 3 Hence proved. 4. Find the acute angles of A and B, if sin (A + 2B) = 3 and cos (A + 4B) = 00, where A > B. 2 [4] Solution: Given that, sin (A + 2B) = 3 2 sin(A + 2B) = sin600 ...(i) A+ 2B = 60° and cos(A + 4B) = 00 cos (A + 4B) = cos 90° ...(ii) A+4B = 90° On subtracting Eq. (i) from Eq. (ii), we get www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 13 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes A 4B 90 0 A 2B 600 2B 300 B = 15° On putting B = 15° in Eq. (i) we get A + 2(15°) = 60° A =60° – 30° = 30° 5. If (cot + tan ) = m and (sec – cos ) = n, then prove that: mn 2/3 mn 2/3 1. [4] Solution: To prove, (m n)2/3 – (mn )2/3 =1 Given, (cot + tan ) = m and (sec – cos ) = n Now, m2n= (cot tan )2 (sec cos ) 2 cos sin 1 = cos sin cos cos 2 cos 1 cos2 = sin cos cos 1 sin2 = . [ cos2 sin2 1] 2 2 sin cos cos 1 sec3 m2n 3 cos ..(i) and mn2 cot tan sec cos 2 2 cos sin 1 = cos sin cos cos 2 cos2 sin2 1 cos2 = sin cos cos sin2 1 = sin cos cos2 2 [ sin2 cos2 1 ] sin4 sin3 = mn2 tan3 3 3 sin cos cos From Eq. (i), sec3 = m2n sec = (m2n)1/3 …(ii) [taking cube root both sides] ..(iii) www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 14 Pioneer Education {The Best Way To Success} tan3 6. IIT – JEE /AIPMT/NTSE/Olympiads Classes From Eq. (ii), = [taking cube root both sides] ...(iv) sec = (m2n )1/3 On squaring Eqs. (iii) and (iv) and then subtracting, we get sec+ – tan2 = (m2n)2/3 – (mn2)2/3 [ sec2 – tan2 = 1] 1 = (m2n)2/3 –(mn2)2/3 Hence proved. tan cot 1 cot tan . Prove that : 1 cot 1 tan Solution: tan cot To prove, 1 cot 1 tan = tan 1 1 tan LHS = = = mn2 [4] 1 tan 1 cot tan 1 tan tan cot 1 cot 1 tan tan 1 1 tan 1 tan 1 tan cot 1 tan tan 1 tan 1 tan 1 tan tan tan2 1 tan3 1 = tan 1 tan tan 1 tan tan 1 tan 1 tan2 tan 1 = tan tan 1 [ a3 – b3 = (a – b) (a2 + ab + b2)] tan2 tan 1 tan2 tan 1 = tan tan tan tan = tan 1 cot 1 cot tan LHS = RHS Hence proved. www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 15 Pioneer Education {The Best Way To Success} 7. Prove that : IIT – JEE /AIPMT/NTSE/Olympiads Classes 1 1 1 1 . cosec cot sin sin cosec cot [4] Solution: To prove, 1 1 1 1 cosec cot sin sin cosec cot 1 1 LHS = cosec cot sin = = cosec cot cosec 1 cosec cot cosec cot cosec cot cosec a ba b a cosec cot cosec cot cosec = 1 = cosec cot cosec 2 2 [ 1 cosec sin 2 b2 cosec2 cot 2 1 ] = cot RHS = 1 1 sin cosec cot = cosec = cosec = cosec cosec cot 1 cosec cot cosec cot cosec cot cosec 2 cot 2 cosec cot 1 = cosec cosec cot [ cosec2 cot 2 1 ] = cot LHS = RHS Hence proved. 1 1 8. If cos and tan then find, Sin where and are both acute angles. [4] 2 3 Solution: 1 1 0 cos 60 Here, cos cos 60° 2 2 a =60° www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 16 Pioneer Education {The Best Way To Success} and 1 3 tan = tan 30° IIT – JEE /AIPMT/NTSE/Olympiads Classes 1 3 tan 300 300 sin sin 600 300 = sin 90° = 1 [ Hence, sin ( ) is 1. sin 90° = 1] 9. Without using trigonometric tables evaluate the following cosec2 900 tan2 2tan2 300 sec2 520 sin2 380 4 cos2 480 cos2 420 cosec2700 tan2 200 Solution: Now, = cosec2 900 tan2 4 cos2 480 cos2 420 4[{cos 90 42 } cos [4] 2tan2 300 sec2 520 sin2 380 cosec2 700 tan2 200 [{cosec 900 }2 tan2 ] 0 2 0 2 2 420 ] 1 0 0 2 2 0 2 . [sec 90 38 } . sin 38 3 {cosec2 900 200 }2 tan2 200 2 .cosec2380. sin 2 38 0 sec tan 3 2 0 = 2 0 2 0 4{sin 42 cos 42 } sec 20 tan2 200 2 2 cosec 900 sec 0 cos 90 sec and sec 900 cosec = 1 2 1 . 2 0 . sin2 380 4 3 sin 38 sec2 tan2 1 2 2 sin cos 1 1 and cosec sin www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 17 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes 1 2 1 . 4 3 1 3 8 5 = 12 12 = 10. Prove that: 1 cot A tan A sin A cos A = sin 2 A cos 2 A. sec 3 A cosec3A Solution: To prove, 1 cot tan A sin A cos A sec 3 LHS = A cosec A 3 [4] sin2 Acos2 A 1 cot A tan A sin A cos A sec 3 A cosec3A cos A sin A 1 sin A cos A sin A cos A = 1 1 cos3 A sin3 A sin A cos A tan A , cot A cos A sin A and sec A 1 , cosec A 1 cos A sin A cos2 A sin2 A 1 sin A cos A sin A cos A = sin3 A cos3 A sin3 A cos3 A 1 1 sin A cos A sin A cos A = 3 sin A cos3 A sin3 Acos3 A 3 3 sin A cos A 1 sin Acos A sin A cos A [sin cos 1] sin3 A cos3 A sin A cos A 2 2 sin A cos A 1 = sin A cos A [ sin 3 × A cos3 A sin A cos A sin A cos A sin2 A cos2 A sin Acos A a3 – b3 = (a – b) (a2 + b2 + ab)] www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 18 Pioneer Education {The Best Way To Success} = = sin A cos A 1 sin 2 2 Acos A IIT – JEE /AIPMT/NTSE/Olympiads Classes sin2 A cos2 A sin A cos A sin A cos A 1 sin2 Acos2 A 1 sin A cos A = sin2 A cos2 A LHS = RHS [ sin2A + cos2 A = 1] Hence proved. www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 19