A SQUARE SERIES TRIGONOMETRY SSC TRIGONOMETRY. [Pick the date]

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1 SSC TRIGONOMETRY A SQUARE SERIES [Pick the date] SSC TRIGONOMETRY A Squre Study Material Arif Baig Cell: , arif4medn@gmail.com, web:

2 . Trigonometry Triogonometry: Trigonometry is the study of relationship of three sides and three angles measures of a triangle. Naming the sides in a right angle triangle: C Hypotenus Opposite side of angle A In triangle ABC CAB (acute angle) AB Adjacent side of angle A BC Opposite side of angle A AC Hypotenus Trigonometric ratios: A B Adjacent side of angle A C SinA CosecA Hypotenus Opposite side of angle A SecA TanA A CosA CotA B Adjacent side of angle A Sine of A SinA BC side to angle A Opposite AC Hypotenus Cosecant of ACosecA AC Hypotenus BC Opposite side to angle A Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

3 Cosine of ACosA AB side to angle A Adjacent AC Hypotenus Secant of ASecA AC Hypotenus AB Adjacent side to angle A Tangent of ATanA BC side to angle A Opposite AB Adjacent side to an gle A Cotangent of ACotA AB side to angle A Opposite BC Adjacent side to angle A SinA and coseca, CosA and SecA, TanA and CotA are reciprocal or multiplicative inverse to each other. SinA CosecA CosA SecA TanA CotA TanA SinA CosA or CosecA SinA or SecA CosA or CotA TanA or CotA CosA SinA Trigonometric Ratios Of Some Specific Angles: Degree Angle A SinA 0 CosA TanA CotA (Not defind) CosecA 0 SecA (Not defind) (Not defind) (Not defind) Note: The values of Sin and Cos always lies between 0 and. Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

4 The values of Tan always lies between 0 and (not defind). The values of Cot always lies between (not defind) and 0. The values of Cosec always lies between (not defind) and. The values of Sec always lies between and (not defind). Trigonometric Ratios Of Complementary Angles: If the sum of two angles is equal to 90 0 then the angles are called complementary angles. In a ABC if B90 0, then sum of other two angles must be 90 o. ( Sum of angles in a triangle 80 o ) i,e; A + C Hence A and C are called complementary angles. If is acute anglethen Sin(90 0 -A)CosA and Cos(90 0 -A)SinA Tan(90 0 -A)CotA and Cot(90 0 -A)TanA Sec(90 0 -A)CosecA and Cosec(90 0 -A)SecA 90 0 A B Trigonometric Identity: Sin A+Cos A Sec A-Tan A Cosec A-Cot A (SinA) Sin A SinA DO THIS Identify Hypotenuse, Opposite side and Adjacent side for the given angles in the given triangles.(page NO.7) C. For angle R P Sol: in the PQR Opposite sidepq Adjacent sideqr HypotenusePR Q R. (i) For angle X Z (ii) For angle Y Sol: in the XYZ i) for angle X Opposite sideyz X Y Adjacent sidexz HypotenuseXY ii) for angle Y Opposite sidexz Adjacent sideyz HypotenuseXY TRY THIS Find lengths of Hypotenuse, Opposite side and c Adjacent side for the given angles in the given triangles.. For angle C 4cm 5cm. For angle A Sol: By Pythagoras theorem B A 4 Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

5 AC AB +BC (5) AB +(4) 5 AB +6 AB 5-6 AB 9 AB ABcm. For angle C: Opposite sideabcm Adjacent sidebc4cm HypotenuseAC5cm For angle C: Opposite sidebc4cm Adjacent sideabcm HypotenuseAC5cm DO THIS. Find (i) sin C (ii) cos C and C (iii) tan C in the adjacent triangle. Sol: By Pythagoras theorem 5cm cm AC AB +BC () AB +(5) 69 AB +5 B A AB 69-5 AB 44 AB ABcm. i) SinC 0pp 0site side to c hypotenuse AB AC 0pp 0site side to c iii) TanC Adjacent side to c AB BC 5 t side to c, ii) CosCAdjacen BC 5, hypotenuse AC. In a triangle XYZ, Y is right angle, XZ 7 m and YZ 5 cm, then find (i) sin X (ii) cos Y (iii) tan X Sol:Given In XYZ, Y is right angle Z XY7cm, YZ5cm By Pythagoras theorem XZ XY +YZ 5cm 7cm (7) XY +(5) 89 XY +5 XY 89-5 Y X XY 64 XY 8 XY8cm. 0pp 0site side to X i) SinX YZ hypot enuse XZ 5 side to Z, ii) CosZAdjacent YZ 7 hypotenuse XZ 5 7 0pp 0site side to X iii) TanX Adjacent side to X YZ XY 5 8. In a triangle PQR with right angle at Q, the value of angle P is x, PQ 7 cm and QR 4 cm, then find sin x and cos x. Sol:Given In a triangle PQR with right angle at Q R angle P is x, PQ 7 cm and QR 4 cm By Pythagoras theorem PR PQ +QR 4cm PR (7) +(4) PR X PR 65 Q 7cm P PR 5 PR5cm. 0pposite side to x Sinx QR hypotenuse PR 4 side to x, CosxAdjacent PQ 5 hypotenuse PR 5 5 Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

6 TRY THESE In a right angle triangle ABC, right angle is at C. BC + CA cm and BC - CA 7cm, then find sin A and tan B. Sol:Given In a triangle ABC with right angle at C And BC+CAcm (), BC-CA7cm..() From () and () we get BC+CA B BC - CA 7 BC0 BC5cm Substituting BC5cm in () 5cm BC+CA 5+CA CA-58cm. By Pythagoras theorem C A AB AC +BC 8cm AB (8) +(5) AB 64+5 AB 89 AB 7 AB7cm. 0pp 0site side to A SinA BC hypotenuse AB 5, 7 0pp 0site side to B TanB Adjacent side to B AC BC 5 THINK AND DISCUSS Discuss between your friends that (i) sin x 4 for some value of angle x. Sol: the value of sin is always lies between o and. But here sin x 4 > so, its does not exist. (ii) The value of sin A and cos A is always less than. Why? Sol: (iii) tan A is product of tan and A. Sol: tana is an abbreviated form of tan of angle A.so, tana is not a product of tan and A. if tan separated from A has no meaning because of here A is angle. TRY THIS What will be the side ratios for sec A and cot A? Hypotenus side to A Sol: SecA, CotAAdjacent Adjacent side to A 0pp 0site side to A THINK AND DISCUSS.is sina is equql to tana? Sol: yes, sina tana opposite side side sina, adjacent hypotenuse hypotenuse.is is equql to cota? sina Sol: yes, cota sina opposite side sina hypotenuse side, adjacent hypotenuse opposite side sina h ypotenuse adjacent side h ypotenuse opposite side tana. adjacent side adjacent side h ypotenuse sina opposite side h ypotenuse adjacent side cota opposite side EXERCISE -.. In right angle triangle ABC, 8 cm, 5 cm and 7 cm are the lengths of AB, BC and CA 6 Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

7 respectively. Then, find out sin A, cos A and tan A. sol: Given the length of the sides of triangle AB8cm, BC5cm and AC7cm We have 7cm 8cm sina BC AC 5, 7 coasaab 8, AC 7 tanabc 5cm. The sides of a right angle triangle PQR are PQ 7 cm, QR 5cm and RP 4 cm respectively. Then find, tan Q -tan R. Sol: Given in PQR PQ7cm, QR5cm and RP4cm R We have tanq RP 4 5cm PQ 7 4cm tanr QP PR 7 4 P tanq-tanr Q B AB cm. In a right angle triangle ABC with right angle at B, in which a 4 units, b 5 units and BAC. Then, find cos and tan. Sol: Given in ABC right angle at B. a4units, b5units, BAC A we have by Pythagoreans theorem 5 AC AB + BC (5) AB + (4) C 4 B AB AB 49AB 7 AB 7cm AB. Hence cos AB AC If cos A tan BC AB 5 7, then find sin A and tan A. C cos A we have by Pythagoreans theorm AC AB + BC () () + BC B A 6944+BC 69-44BC 5BC 5 BC 5cmBC We have sina BC AC 5, BC AB 5 A C 5. If tan A 4, then find sin A and cos A. tan A 4 C tana 4 hence, 4 by Pythagoreans theorm AC AB + BC AC + 4 B A AC 9+6 AC 5 AC 5 AC5cm We have sina BC AC 4 5, AB AC If A and X are acute angles such that cos A cos X then show that A X. Sol: Let ABC, PQR X Given cosx We have AB AC cosxxy XZ Then AB C A XY AC XZ by Pythagoreans theorm AC AB + BC and XZ XY + YZ AC AB + BC and XZ XY + YZ So, AC AB BC XZ XY YZ A X. 7. Given cot 7 (+sin )( sin ), then evaluate (i) 8 (+cos )( cos ) (ii) +sin cos Sol: Given cot 7 A 8 by Pythagoreans theorm AC AB + BC 8 AC AC B 7 C AC AC. We have sin AB BC 8, cosbc AC 7 (i) (+sin )( sin ) (+cos )( cos ) () sin () cos ( sin ) ( cos ) cos cos sin cot ( 7 8 ) (ii) +sin cos cos sin sin 7 Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

8 In a right angle triangle ABC, right angle is at B, if tan A then find the value of (i) sin A cos C + cos A sin C (ii) cos A cos C - sin A sin C a right angle triangle ABC, right angle is at B, tan A so B90 0, tan A tan A tan60 0 A60 0 We know in a triangle A+B+C C C80 0 C (i) sin A cos C + cos A sin C sin60 0 cos0 0 + cos60 0 sin (ii) cos A cos C - sin A sin C cos60 0 cos0 0 sin60 0 sin Exercise:.. Evaluate the following (i) sin45 0 +cos45 0 Sol: Given sin45 0 +cos (ii) cos 45 0 sec 0 0 +cosec 60 0 Sol: Given cos 45 0 sec 0 0 +cosec (iii) sin 00 +tan 45 0 cos ec60 0 cot45 0 +cos 60 0 sec 0 0 Sol: Given + 4 sin 00 +tan 45 0 cos ec cot cos 60 0 sec (iv) tan cos 0 0 sin 60 0 Sol: Given tan cos 0 0 sin () +( ) ( ).(). (v) sec 60 0 tan 60 0 sin 0 0 +cos 0 0 Sol: Given sec 60 0 tan 60 0 sin 0 0 +cos 0 0() ( ) ( ) +( 4 ) Choose the right opinion and justify your choice. (i) tan 0 0 +tan 45 0 (a) sin60 0 (b)cos60 0 (c) tan0 0 (d) sin0 0 tan 0 0 +tan 45 tan0 0 (ii) tan tan () +. (a) tan90 0 (b) (c) sin45 0 (d) 0 tan tan 45 0 () +() Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

9 (iii) tan 0 0 tan 0 0 (a) cos60 0 (b)sin60 0 (c) tan60 0 (d) sin0 0 tan 00 tan 0 tan ( )... Evaluate sin60 0.cos0 0 +sin0 0.cos60 0. what is the value of sin( ), what can you conclude. sin60 0.cos0 0 +sin0 0.cos sin( ) sin90 0. We know that in PQR 0 0 tan60 0 QR PQ QR 6 6 QR sin60 0 QR PR 6 PR PRcm. Q 6cm P 6. In XYZ, right angle at Y, YXx and XZx then determine YXZ and YZX. Sol: Given In XYZ, right angle at Y, YZx and XYx By Pythagoreans theorem sin60 0.cos0 0 +sin0 0.cos60 0 sin( ) XZ XY +YZ Z We conclude that (x) (x) +YZ x sina.cosb+sina.cosb sin(a + B). 4.Is it right to say cos( ) cos60 0.cos0 0 +sin60 0.sin0 0 cos( ) cos60 0.cos0 0 +sin60 0.sin0 0 cos x -x YZ Y x X x YZ x YZ xyz We know In XYZ tanx YZ XY X60 0. x x tan60 0 Yes, it is right to say cos( ) cos60 0.cos0 0 +sin60 0.sin In right angle triangle PQR right angle is at Q and PQ6cm RPQ Determine the length of QR and PR. tanz XY YZ Z0 0. x x tan00 Hence YXZ60 0, YZX0 0. Sol: given PQ6cm and RPQ 60 0 R 7. Is it right to say that sin(a+b)sina+sinb justify your answer. Sol: Given sin(a+b)sina+sinb 9 Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

10 Let A0 0, B60 0 Sin( ) sin0 0 + sin60 0 sin Isn t right to say sin(a+b)sina+sinb.. Evaluate tan 60 (i) cot 54 0 tan 60 cot 54 0 cotatan(90 0 A) Exercise:. cot6 0 tan( )tan6 0 tan 60 cot 6 0. (ii) cos 0 sin78 0 cos 0 sin78 0 sin(90 0 A) cos 0 sin( )sin78 0 sin78 0 sin78 0. (iii) cosec 0 sec59 0 cosec 0 sec59 0 cosecasec(90 0 A) cosec 0 sec( )sec59 0 sec59 0 sec59 0 cotatan(90 0 A) cot6 0 tan( )tan6 0. (iv) sin5 0. sec75 0 sin5 0. sec75 0 sin 50 cos 75 0 sin(90 0 A) cos75 0 sin( )sin5 0 sin50 sin 5 0. (v) tan6 0. tan64 0 tan6 0. tan64 0 tanacot(90 0 A) tan64 0 cot( )cot6 0 tan6 0. cot6 0 tan6 0.. tan 6 0. (i) Show that tan48 0. tan6 0. tan4 0. tan74 0 Sol: LHS. tan48 0. tan6 0. tan4 0. tan74 0 tanacot(90 0 A) tan4 0 cot( )cot48 0 tan74 0 cot( )cot6 0 tan48 0. tan6 0. cot48 0. cot6 0 tan48 0. tan6 0.. RHS. tan 48 0 tan 6 0 (ii) cos6 0.cos54 0 -sin6 0.sin Sol: LHS cos6 0.cos54 0 sin6 0.sin54 0 sin(90 0 A) 0 Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

11 cos6 0 sin( )sin54 0 cos54 0 sin( )sin6 0 sin54 0.sin6 0 sin6 0. sin54 0. RHS. If tanacot(a 8 0 ), where A is an acute angle. Find the value of A. tanacot(a 8 0 ) tanacot(90 0 A) tanacot(90 0 A) cot(90 0 A) cot(a 8 0 ) 90 0 A A A+A 08 0 A 6 0 A. 4. If tana cotb where A and B are acute angles prove that A+B90 0 tana cotb cotatan(90 0 A) cotbtan(90 0 B) tana tan(90 0 B) A + B 90 0 C LHS. tan( A+B ) tanacot(900 A)cotA tan(90 0 C ) cot C RHS. 6. Express sin75 0 +cos65 0 in terms of trigonometric ratios of angle between 0 0 and sin75 0 +cos65 0 sinacos(90 0 A) cos75 0 sin( )sin5 0 sin(90 0 A) cos65 0 sin( )sin5 0 cos5 0 +sin5 0..Evaluate the following Exercise:.4 (i) (+tan -sec )(+cot -cosec ) (+tan -sec )(+cot - cosec ) (+ sin cos cos )(+cos ) sin sin cos +sin cos sin+cos sin A90 0 B A+B If A,B and C are the interior angles of a triangle ABC then show that tan( A+B ) cot C. Sol: We know that in triangle ABC A+B+C 80 0 (dividing the equation by ) (cos +sin ) () cos.sin cos +sin +cos.sin cos.sin +cos.sin cos.sin cos.sin cos.sin. sin +cos A+B+C 800 A+B +C 900 (ii) (sin + cos) + (sin cos) sol: (sin + cos) + (sin cos) sin +cos +sin.cos +sin +cos -sin.cos Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

12 +. (iii) ( sec )( cosec -) Sol: ( sec )( cosec -) tan sec - cot cosec - tan. cot tan.. tan. Show that ( cosec cot ) cos +cos seca+tana. RHS 4.Show that tan A cot A tan A. Sol: LHS. tan A cot A tan A tan A tan A tan A tan A tan A tan A. RHS. tan A tan A Sol: LHS. 5. Show that cos cos tan.sin ( cosec cot ) sin cos sin Sol: LHS. cos cos cos cos sin cos cos sin sin cos ( cos ) sin sin cos.sin ( cos ) cos ( cos )( cos ) ( cos )(+cos ) ( cos ) (+cos ). RHS tan.sin RHS. 6. Simplify seca(-sina)(seca+tana). Sol: seca(-sina)(seca+tana) (seca-seca.sina)(seca+tana).show that +sina sina seca+tana (seca- sina )(seca+tana) Sol: LHS. +sina sina (seca-tana)(seca+tana) sec A tan A sec A tan A Multiply and divided by (+sina). +sina +sina sina +sina (+sina ) sina (+sina ) 7. Prove that (sina + coseca) + ( + seca)7+tana+cota. Sol: LHS. (sina + coseca) + ( + seca) sin A + coseca +.sina.coseca+cos A + seca +..seca +sina + sina seca tana sina sin A+cos A+cosecA +seca cota + + tana +4 coseca + cota seca + tana Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web:

13 7+tanA + cota. RHS. 8. Simplify (-cos )(+cos )(+cot ) cos cos k k + (k +) 4k (k +) (-cos )(+cos )(+cot ) ( cos )( + cot ) sin (+ cos sin ) sin ( sin +cos ) sin + cos sin sin + cos. 9. sec +tan p then what is the value of sec - tan? sec +tan p sec -tan sec +tan cos cos (k ) (k +) (k ) (k +) RHS. sec -tan p 0. If cosec +cot k then prove that cos k k + cosec +cot k cosec -cot cosec +cot cosec -cot k From and cosec +cot k cosec -cot k cosec k+ k cosec k + k cosec k + k cos - sin cos sin Prepared By:ARIF BAIG Cell: , arif4medn@gmail.com., Web: