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Simple Problems In Mathematics

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This presentation is about Simple Problems in Mathematics.

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    Trigonometry THE HALL
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    SOME PROBLEMS 1. Prove that Cosec 9/ (Cosec e - 1) + Cosec 9/ (Cosec e + 1) - 2 LHS - {Cosec em(Cosec e + 1) + Cosec em (Cosec e -1) l/ (Cosec e e + 1) - {Cosec2 9+ Cosec 9 + Cosec 29- Cosec 9) / (Cosec 29 -I) 2 Cosec29/ (Cosec 29 -I) 2 Cosec29/ COt29 (2/ Sin29)/ (COS2e / Sin29) - 2/ cos2e — 2 Sec2e - RI-IS (Since, cosec2e - 1 + COt29)
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    2. A. If tan29 - I-m2, prove that sec 9 + tan 39ecosec 9 - +/- (2-m2)3/2 We have tan2e = I-m2 Now, Sec2e = 1 + tan2e = 2 — m2 Hence, Sec e = Now, LHS sec e + tan 36.cosec 9 = sece + tan2e ( sin 9/cos e . l/sin 9) = sec e + tan2e. sec 9 = sec e (1+ tan29) = sec Sec 29 = + /- (2-m2)3/2 - RHS PROVED
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    3. If e lfies in the fourth quadrant, and Cos 9 - 5/13 , find the values of i. O Sn 9 and tan e (13 Sin e + 5 Sec (5 tan e + 6 Cosec 0) 5 13 Let the fig. be drawn, as given cos OM/OP = 5/13 Hence, 0M = 5k, OP - 13k By Pythagoras Theorem 0M2 + MP2 op2 or, (5k)2 + MP2 - (13k)2 or, MP2 = (13k)2- (5k)2 p 144k2 Or, MP = +-12k As e lies in the 4th. Quadrant, MP is -ve, and = -12k
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    о 5 13 м 12 р llE Sin е = МР/ОР= -12/13 Тап е = МУОМ - -12/5 Also, Sec е = ОР/ОМ = 13/5 Cosec е = ОР/МР = -13/12 (13 Sin + 5 Sec tan + 6 Cosec е) = {134-12/13) + 5 в1З/5И5.(-12/5) + 64- 13/12) = - 13/2) = -2/37
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    4. Fmn&'XTfromtheeqUätlon Cosec (900-9) - x (900-9) tan (180+9) - Cosec (900-9) - Sec e Sin (900-6) = cos e tan (1800+9) = tan e Sin (900+9) — cos e Sin (900+9) Hence, Sec 6 - x Cos e tan 9 = Cos 9 Or, xSine -Sece- Cos 9 - (1- cos2evcos e = Sin29/ Cos e Or, x = Sin 9/Cos e Or, x = tan e
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    5. Prove ha = (tan A + tan B)/(l - tan A tan B) tan B) tan (A+B) = Sin (A+B)/Cos (A+B) = (Sin A cos B + cos A Sin (Cos A cos B -Sin B) Dividing numerator and denominator by Cos A Cos B, we get = A cos B + cos A B)/Cos A cos (Cos A cos B - Sin A Sin B)/Cos A cos B)) = (tan A + tan tan A tan B) PROVED
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    aaĄoud (O + x) ułs (O SHU - (o — e) (o + x) + (o — e) soo — — e) '(0 + x) soo + (O — e) soo (O + x) ułs} — (O ) ułs/(e ) WS — — e) soo (O ) ułs/(e + x )
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    If CosA - 1/7, and Cos B - 13/14 (A, B being positive acute), then prove that A CosA - 1/7 Sin A ( 1 — cos 2 A) — ( 48/49) 443/7 Also, cos B- 13/14 B - cos 2B) - 169/196) - ( 27/196) = 3M3/14 cos (A-B) - CosA cos B + Sin A Sin B - (1/7), (13/14) + (4 M3/7)E (3M3/14) - (13 + 36)/98 - 49/98 - - cos 600 Hence, B - 600 — B 600
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    8. If the angles A, B, C are in AP, prove that Cot B - (Sin A - (Cos C - cos A) RI-IS = (Sin A - (Cos C - cos A) = (2 Sin (A-C)/2.Cos {2 = {Cos - Cot (A+C)/2 Cot B ( smnce, A, B, Care in AP; (A+C)/2 - B) - LHS PROVED
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    Thank You

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