f=dks.kferh; vuqikr (Trigonometric Ratios) PDF

f=dks.kferh; vuqikr Chapter f=dks.kferh; vuqikr 03 (Trigonometric Ratios) f=dks.kferh; vuqikr dks f=dks.kfefr pj.k&I ds :i esa Hkh tkuk tkrk gSA CONTE...

f=dks.kferh; vuqikr Chapter f=dks.kferh; vuqikr 03 (Trigonometric Ratios) f=dks.kferh; vuqikr dks f=dks.kfefr pj.k&I ds :i esa Hkh tkuk tkrk gSA CONTENT 1- ifjp; (INTRODUCTION) 1. ifjp; 'f=dks.kfefr' 'kCn dh O;qRifÙk xzhd 'kCnks 'fVªxksu' vkSj 'esVªksu' ls gqbZ gS rFkk 2. vk/kkjHkwr f=dks.kferh; bldk vFkZ 'f=Hkqt dh Hkqtkvksa vkSj dks.kksa dks ekiuk' gksrk gSA loZlfedk,¡ dks.k: 3. egRoiw.kZ f=dks.kferh; ,d dks.k og eki gS tks ,d fdj.k ds mlds çkjafHkd fcanq ds ifjr% ?kw.kZu ij vuqikr curk gSA ewy fdj.k dks çkjafHkd Hkqtk dgk tkrk gS vkSj ?kw.kZu ds ckn fdj.k 4. lac) dks.kksa ds f=dks.kferh; dh vafre fLFkfr dks dks.k dh vafre Hkqtk dgk tkrk gSA ?kw.kZu fcanq dks 'kh"kZ Qyu dgrs gSaA ;fn ?kw.kZu dh fn'kk okekorZ gS] rks dks.k dks /kukRed dgk tkrk gS 5. f=dks.kferh; Qyu vkSj ;fn ?kw.kZu dh fn'kk nf{k.kkorZ gS] rks dks.k _.kkRed gksrk gSA 1. nks dks.kksa ds ;ksx ;k varj ds f=dks.kferh; Qyu izkjafHkd Hkqtk 6. :ikUrj.k lw= 'kh"kZ 0 A 7. vioR;Z dks.k vkSj viorZd dks.k vfUre Hkqtk 8. rhu dks.k ¼okekorZ eki½ ¼ii½_.kkRed dks.k ¼nf{k.kkorZ eki½ B 9. çfrca/k loZlfedk,¡ dks.kksa ds ekiu ds fy, ç.kkfy;k¡: 10. SINE vkSj COSINE Js.kh fuEufyf[kr ç.kkfy;ksa esa dks.k dks ekik tk ldrk gSA 11. COSINE dks.kksa dh xq.kuQy (1) "kkf"Vd i}fr (fczfV'k i}fr): Js.kh 12. f=dks.kfefr; Qyuksa ds 1 bl ç.kkyh esa ,d iw.kZ o`Ùkkdkj /kqeko ds Hkkx dks fMxzh (°) dgk vf/kdre vkSj U;wure eku 360 tkrk gS] ,d fMxzh dss 1 Hkkx dks ,d feuV (’) dgk tkrk gS rFkk 60 feuV ds 1 Hkkx dks lsdaM () dgk tkrk gS 60 ledks.k = 90°, 1° = 60, 1 = 60 (2) 'kfrd i}fr (Ýsp a i}fr): 1 1 bl ç.kkyh esa ,d iw.kZ o`Ùkkdkj /kqekoks ds Hkkx dks xzsM (g) dgk tkrk gS] ,d xzsM ds Hkkx dks ,d 400 100 1 feuV (‘) dgk tkrk gS vkSj ,d feuV ds Hkkx dks lsdM a (“) dgk tkrk gSA 100  ledks.k = 100g; 1g = 100‘; 1‘ = 100“ DETECTIVE MIND "kkf"Vd i}fr esa feuV vkSj lsdM a ] 'kfrd i}fr esa Øe'k% feuV vkSj lsdaM ls fHkUu gksrs gSaA nksuksa i}fr;ksa esa çrhd Hkh fHkUu gksrs gSaA xf.kr (3) jsfM;u ;k o`Ùkh; i}fr ,d o`Ùk ds pki }kjk varfjr dks.k ftldh yackbZ o`Ùk ds dsaæ ij o`Ùk dh f=T;k ds cjkcj gksrh gS] jsfM;u dgykrk gSA bl ç.kkyh esa ekiu dh bdkbZ jsfM;u ( c ) gSA pw¡fd bdkbZ f=T;k okys ,d o`Ùk dh ifjf/k 2 gS] blfy, çkjafHkd Hkqtk dk ,d iw.kZ ifjØe.k 2- jsfM;u dk dks.k cukrk gS blfy,] f=T;k r ds ,d o`Ùk es]a yackbZ r dk ,d pki 1 jsfM;u dk dks.k varfjr djsxkA ge tkurs gSS fd o`Ùk ds leku pki] dsæa ij leku dks.k varfjr djrs gSaA ;fn f=T;k r okys ,d o`Ùk esa] yackbZ  dk ,d pki ,d dks.k varfjr djrk gS ftldk eki 1 jsfM;u gksrk gS] yackbZ  dk ,d pki ,d dks.k varfjr djsxk ftldk eki jsfM;u gS bl çdkj] ;fn f=T;k r ds ,d o`Ùk esa] yackbZ  dk pki dsUnz ij ,d dks.k  jsfM;u r cukrk gS rc  = ;k = r gSA r jsfM;u jsfM;u jsfM;u jsfM;u DETECTIVE MIND ;fn dks.k dk eki fn[kkrs le; fdlh izrhd dk mYys[k ugha fd;k x;k gks] rks bls jsfM;u esa ekik tkrk gSA tSls  = 15 dk vFkZ 15 jsfM;u gksxkA o`Ùk[k.M dk {ks=Qy : {ks=Qy = 1 r2 oxZ bdkbZ;k¡ 2 jsfM;u] fMxzh vkSj xzsM ds chp laca/k :  jsfM;u = 90° = 100g 2 U;wu dks.kksa ds fy, f=dks.kferh; vuqikr : eku yhft, fd ,d ifjØkeh fdj.k OP] OA ls izkjaHk gksrh gS vkSj OP dh fLFkfr esa ?kwerh gS] bl çdkj dks.k AOP cukrh gSA ifjØkeh fdj.k esa dksbZ fcanq P yhft, vkSj çkjafHkd fdj.k OA ij yac PM [khph,¡A ledks.k f=Hkqt MOP esa] OP d.kZ gS] PM yac gS] vkSj OM vk/kkj gSA dks.k AOP ds f=dks.kferh; vuqikr ;k Qyu fuEukuqlkj ifjHkkf"kr fd, x, gSa: MP yEc ] vFkkZr] ] dks.k AOP dk Sine dgykrk gSA OP d.kZ OM vk/kkj ] vFkkZr] ] dks.k AOP dk Cosine dgykrk gSA OP d.kZ MP yEc ] vFkkZr] ] dks.k AOP dk Tangent dgykrk gSA OM vk/kkj OM vk/kkj ] vFkkZr] ] dks.k AOP dk Cotangent dgykrk gSA MP yEc OP d.kZ ] vFkkZr] ] dks.k AOP dk Secant dgykrk gSA OM vk/kkj OP d.kZ ] vFkkZr] ] dks.k AOP dk Cosecant dgykrk gSA MP yEc  f=dks.kferh; vuqikr 2. vk/kkjHkwr f=dks.kferh; loZlfedk,¡ (BASIC TRIGONOMETRIC IDENTITIES) : (1) sin2 + cos2 = 1 ; −1  sin   1; −1  cos   1    R (2) sec2 − tan2 = 1 ; sec   1    R (3) cosec2 − cot2 = 1 ; cosec   1    R 3. egRoiw.kZ f=dks.kferh; vuqikr (IMPORTANT TRIGONOMETRIC RATIOS) : (1) sin n  = 0 ; cos n  = (-1)n ; tan n  = 0 tgk¡ n  I gSaA (2n + 1)  (2) sin = (−1)n vkSj cos (2n + 1)  = 0 tgk¡ n  I gSaA 2 2  3 −1 5 (3) sin 15° ;k sin = = cos 75° ;k cos ; 12 2 2 12  3 +1 5 cos 15° ;k cos = = sin 75° ;k sin ; 12 2 2 12 3 −1 3 +1 tan 15° = = 2 − 3 = cot 75° ; tan 75° = = 2 + 3 = cot 15° 3 +1 3 −1  2− 2  2+ 2  3 (4) sin = ; cos = ; tan = 2 − 1 ; tan = 2 +1 8 2 8 2 8 8  5 −1 5 +1 (5) sin ;k sin 18° = = cos72° vkSj cos 36° ;k cos  = = sin54° 10 4 5 4 4. lac) dks.kksa ds f=dks.kferh; Qyu (TRIGONOMETRIC FUNCTIONS OF ALLIED ANGLES): ;fn  dksbZ dks.k gS] rks − ] 90 ± ] 180 ± ] 270 ± ] 360 ±  vkfn lac) dks.k dgykrs gSaA (1) sin (− ) = − sin  ; cos (− ) = cos  (2) sin (90°- ) = cos  ; cos (90° − ) = sin  (3) sin (90°+ ) = cos  ; cos (90°+ ) = − sin  (4) sin (180°− ) = sin  ; cos (180°− ) = − cos  (5) sin (180°+ ) = − sin  ; cos (180°+ ) = − cos  (6) sin (270°− ) = − cos  ; cos (270°− ) = − sin  (7) sin (270°+ ) = − cos  ; cos (270°+ ) = sin  5. f=dks.kferh; Qyu (TRIGONOMETRIC FUNCTIONS): (i) y = sin x izkra : x  R ifjlj: y  [–1, 1] y 1 0 3    3 x –2 – 2 2 2 2 2 –1 (ii) y = cos x izkra : x  R ifjlj: y  [ – 1, 1] y 1 0 3    3 x –2 – 2 2 2 2 2 –1 xf.kr  (iii) y = tan x izkra : x  R – (2n + 1)  , n    2 ifjlj: y  R y   – 2 0 2  x – 32 3 2 (iv) y = cot x izkra : x  R – {n}, n   ifjlj: y  R y –    3 2 x 0 2 2 2 (v) y = cosec x izkra : x  R – {n}, n   ifjlj : y  (− , − 1]  [1, )  (vi) y = sec x izkra : x  R – (2n + 1)  , n    2 ifjlj : y  (− , − 1]  [1, ) y 1 – – 0   3 2 x 2 2 2 –1 6. nks dks.kksa ds ;ksx ;k varj ds f=dks.kferh; Qyu (TRIGONOMETRIC FUNCTIONS OF SUM OR DIFFERENCE OF TWO ANGLES) : (1) sin (A ± B) = sinA cosB ± cosA sinB (2) cos (A ± B) = cosA cosB sinA sinB (3) sin²A − sin²B = cos²B − cos²A = sin (A+B). sin (A− B) (4) cos²A − sin²B = cos²B − sin²A = cos (A+B). cos (A − B) tanA  tanB (5) tan (A ± B) = 1 tanA tanB cotAcotB 1 (6) cot (A ± B) = cotB  cotA  f=dks.kferh; vuqikr 7. :ikUrj.k lw= (TRANSFORMATION FORMULAE) : C+D C −D (i) sin(A+B) + sin(A − B) = 2 sinA cosB (v) sinC + sinD = 2 sin cos 2 2 C+D C −D (ii) sin(A+B) − sin(A − B) = 2 cosA sinB (vi) sinC − sinD = 2 cos sin 2 2 C+D C −D (iii) cos(A+B) + cos(A − B) = 2 cosA cosB (vii) cosC + cosD = 2 cos cos 2 2 C+D D −C (iv) cos(A − B) − cos(A+B) = 2 sinA sinB (viii) cosC − cosD = 2 sin sin 2 2 SOLVED EXAMPLES sin2 y 1 + cosy siny Example: 1 O;tad 1 – + – dk eku cjkcj gS & 1 + cosy siny 1 − cosy (1) 0 (2) 1 (3) sin y (4) cos y 1 + cosy 1 + cosy − sin2 y 1 − cos y − sin y cosy + cos2 y 2 2 sin2 y siny Solution: 1– + – = + = + 0 = cos y 1 + cosy siny 1 − cosy 1 + cosy siny (1 − cosy) 1 + cosy Example: 2 ;fn cosec  – sin  = m vkSj sec  – cos  = n gks] rks (m2n)2/3 + (n2m)2/3 cjkcj gSA (1) 0 (2) 1 (3) &1 (4) 2 Solution: cosec  – sin  = m 1 cos2  m= – sin  =...(i) sin sin  1 sin2  n= – cos  =...(ii) cos cos  Lkeh- (i) vkSj (ii) ls cos2  sin2  m×n=. = sin  cos  sin  cos  Lkeh- (i) ls cos2  = m. sin  ;k cos3  = m sin  cos  = m. (mn) = m2n blh çdkj sin3  = n2m vr: sin2  + cos2  = 1 (n2m)2/3 + (m2n)2/3 = 1 Example: 3 fl) dhft, fd (i) sin (45° + A) cos (45° – B) + cos (45° + A) sin (45° – B) = cos (A – B)    3  (ii) tan  +   tan  +   = –1 4   4  Solution: (i) sin (45° + A) cos (45° – B) + cos (45° + A) sin (45° – B) (sin (A ± B) = sinA cosB ± cosA sinB) = sin (45° + A + 45° – B) = sin (90° + A – B) = cos (A – B)    3  1 + tan −1 + tan (ii) tan  +   × tan  +   = × =–1 4   4  1 − tan 1 + tan  xf.kr fl) dhft, fd cos7A + cos8A = 2cos  15A  A Example: 4  cos   gSA 2 2      15A  A Solution: L.H.S. cos7A + cos8A = 2cos   cos    2  2 C +D C −D [ cos C+ cos D = 2 cos cos ]] 2 2 Example: 5 2sin3 sin – cos2 + cos4 dk eku Kkr dhft,A Solution: 2sin3 sin – cos2 + cos4 = 2 sin 3 sin  – 2sin3 sin = 0 Example: 6 fl) dhft, fd sin8cos  − sin6cos3 (i) = tan 2 cos2cos  − sin3sin4 (ii) ;fn A + B = 45° gks] rks fl) dhft, fd (1 + tanA) (1 + tanB) = 2 gSA 2sin8cos  − 2sin6cos3 sin9 + sin7 − sin9 − sin3 2sin2cos5 Solution: (i) = = = tan 2 2cos2cos  − 2sin3sin4 cos3 + cos  − cos  + cos7 2cos5cos2 (ii) A + B = 45° tanA + tanB tan (A + B) = 1  =1 1 – tanAtanB tanA + tanB = 1 – tanA tanB  tanA + tanB + tanA tanB + 1 = 2 (1 + tanA) (1 + tanB) = 2 Example: 7 2(sin6 + cos6) – 3 ( sin4 + cos4 ) + 1 cjkcj gS (1) 0 (2) 1 (3) &2 (4) buesa ls dksbZ ugha Solution: 2[ (sin2 + cos2 )3 –3 sin2  cos2 ( sin2 + cos2 )] –3[ (sin2 + cos2 )2 ] – 2sin2 cos2 +1 ⇒ 2 [ 1 – 3 sin2  cos2 ] – 3 [ 1 –2 sin2 cos2 ] + 1 ⇒ 2–6 sin2 cos2 – 3 + 6 sin2 cos2 + 1 = 0 Example: 8 ;fn 3 sin  + 4 cos  = 5 gks] rks 4 sin  – 3 cos  dk eku gS (1) 0 (2) 1 (3) 2 (4) 3 Solution: ekuk 4 sin  – 3 cos  = a vc  dks foyqIr djus ds fy, lehdj.kksa 3sin + 4 cos  = 5 vkSj 4 sin  – 3 cos  = a, dks oxZ djus vkSj tksMu+ s ij] (3 sin  + 4 cos )2 + (4 sin  – 3 cos )2 = 25 + a2 9 sin2  + 16 cos2  + 24 sin  cos  + 16 sin2  + 9 cos2  – 24 cos  sin  = 25 + a2 9 + 16 = 25 + a2 ;k a2 = 0 a=0 vr% 4 sin  – 3 cos  = 0 Example: 9 ;fn x = r sin  cos , y = r sin  sin  vkSj z = r cos  gks] rks x2 + y2 + z2 dk eku cjkcj gSA (1) 2r2 (2) r2 (3) 0 (4) buesa ls dksbZ ugha Solution: ;gk¡ x + y + z = r sin  cos2  + r2 sin2. sin2  + r2 cos2 2 2 2 2 2 = r2 sin2 ( cos2  + sin2 ) + r2 cos2  = r2 sin2 + r2 cos2  = r2 ( sin2 + cos2) = r2 x2 + y2 + z2 = r2  f=dks.kferh; vuqikr Example: 10 3 cosec20° – sec 20° cjkcj gS (1) 0 (2) 1 (3) 2 (4) 4  3 1  4  cos20 − sin20  3 cos20 − sin20 2 2 =   3 1 Solution: − = sin20 cos20 sin20.cos20 2sin20 cos20 ( sin60.cos20 − cos60.sin20) = 4. sin 40 sin(60 − 20) sin40 =4 =4. =4 sin40 sin40 Example: 11 sin78° – sin66° – sin42° + sin 6° cjkcj gSA (1) 1/2 (2) –1/2 (3) 2 (4) –2 Solution: O;atd = (sin78° – sin42°) – (sin 66° – sin6°) = 2cos (60°) sin (18°) – 2 cos 36°. sin 30°  5 −1   5 +1  1  4  −  4  = − 2 = sin18° – cos36° =      1 − sin  Example: 12 [–, ] esa  ds lHkh laHkkfor ekuksa dk leqPp; bl çdkj Kkr dhft, fd , ( sec  – tan ) ds 1 + sin  cjkcj gS           (1)  , −  (2)  ,  (3)  − ,  (4) buesa ls dksbZ ugha 2 2  2 2   2 2  Solution: Li"Vr%  / 2 rks sec  – tan  = 1 − sin...(i) cos  (1 − sin ) 2 1 − sin 1 − sin 1 − sin rFkk = = = ….(ii) 1 + sin cos  2 cos  cos  leh- (i) vkSj (ii) ls nks O;atd dsoy rHkh cjkcj gksrs gSa ;fn cos  > 0 gks , vFkkZr – /2 <  <  / 2 gks 1 − sin   vkSj sec  – tan  cjkcj gksrs gSa 1 + sin    dsoy tc    − ,  gksA  2 2   Example: 13 ;fn 1 + sin  +   + 2 cos  −   gks] rks f() dk vf/kdre eku gS A  4  4  (1) 1 (2) 2 (3) 3 (4) 4   Solution: gekjs ikl gS 1 + sin  +   + 2 cos  −   4  4  1  1  =1+ (cos  + sin  ) + 2 ( cos  + sin ) = 1 +  + 2  (cos  + sin ) 2  2   1    =1+  + 2 . 2 cos   −  dk vf/kdre eku  2   4  1  1+ + 2 . 2 = 4 gksxk A  2  xf.kr Example: 14 sin 20° sin 40° sin 60° sin 80° dk eku gSA (1) 3/8 (2) 1/8 (3) 3/16 (4) buesa ls dksbZ ugha Solution: sin 20° sin 40° sin 60° sin 80° 3 = sin20 sin( 60 − 20 ) sin( 60 + 20 ) 2 3  sin20 ( sin2 60 − sin2 20 ) = 3 3 = sin20  − sin2 20  2 2  4  3 3 3 3 3 = (3sin20 − 4sin3 20) = sin60 =. = 8 8 8 2 16 Example: 15 sin600° dk eku gSA 3 Solution: sin( 3(180) + 60) = − sin60 = − 2 8. vioR;Z dks.k vkSj viorZd dks.k (MULTIPLE ANGLES AND HALF ANGLES)   (1) sin 2A = 2 sinA cosA ; sin  = 2 sin cos 2 2 (2) cos2A = cos2A − sin2A = 2cos2A − 1 = 1 − 2 sin2A ;     cos  = cos2 − sin² = 2cos2 − 1 = 1 − 2sin2. 2 2 2 2 1 − cos2A 2 cos2A = 1 + cos 2A , 2sin2A = 1 − cos 2A ; tan2A = 1 + cos2A   2 cos2 = 1 + cos , 2 sin2 = 1 − cos . 2 2 2 tanA 2 tan( 2) (3) tan 2A = ; tan  = 1 − tan2 A 1 − tan2 ( 2) 2 tanA 1 − tan2 A (4) sin 2A = , cos 2A = 1 + tan2 A 1 + tan2 A (5) sin 3A = 3 sinA − 4 sin3A (6) cos 3A = 4 cos3A − 3 cosA 3 tanA − tan3 A (7) tan 3A = 1 − 3 tan2 A 9. rhu dks.k (THREE ANGLES): (i) sin (A + B + C) = sin A cos B cos C + sin B cos A cos C + sin C cos A cos B – sin A sin B sin C (ii) cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C tan A + tanB + tanC − tan A tan B tan C (iii) tan (A + B + C) =. 1 − tanA tan B − tan B tan C − tan C tan A  f=dks.kferh; vuqikr DETECTIVE MIND tan (1 + 2 + 3 +....... + n) = tgk¡ Si ,d le; esa fy, x, dks.kksa i ds tangent ds xq.kuQy ds ;ksx dks n'kkZrk gSA 10. çfrca/k loZlfedk,¡ (CONDITIONAL IDENTITIES) : ;fn A + B + C =  gks] rks (i) sin2A + sin2B + sin2C = 4 sinA sinB sinC A B C (ii) sinA + sinB + sinC = 4 cos cos cos 2 2 2 (iii) cos 2 A + cos 2 B + cos 2 C = − 1 − 4 cos A cos B cos C A B C (iv) cos A + cos B + cos C = 1 + 4 sin sin sin 2 2 2 (v) tanA + tanB + tanC = tanA tanB tanC A B B C C A (vi) tan tan + tan tan + tan tan = 1 2 2 2 2 2 2 A B C A B C (vii) cot + cot + cot = cot. cot. cot 2 2 2 2 2 2 (viii) cot A cot B + cot B cot C + cot C cot A = 1  (ix) A + B + C = gks] rks tan A tan B + tan B tan C + tan C tan A = 1 2 11. Sine vkSj Cosine J`a[kyk (Sine and Cosine Series ) : n sin 2  n−1  (i) sin  + sin ( + ) + sin ( + 2) +...... + sin  + (n − 1)  =  sin   +  sin 2  2  n sin 2  n −1  (ii) cos  + cos ( + ) + cos ( + 2 ) +.... + cos  + (n − 1)  =  cos   +  sin 2  2  tgk¡ :   2m, m   12. Cosine dks.kksa dh xq.kuQy J`a[kyk (PRODUCT SERIES OF COSINE ANGLES) sin2n  cos . cos 2. cos22. cos23...... cos2n–1 = 2n sin 13. f=dks.kferh; Qyuksa ds vf/kdre vkSj U;wure eku (MAXIMUM & MINIMUM VALUES OF TRIGONOMETRIC FUNCTIONS) : (1) a2tan2 + b2cot2  dk U;wure eku 2ab gS tgk¡   R gSA (2) acos + bsin dk vf/kdre vkSj U;wure eku a + b rFkk – a + b gSA 2 2 2 2 (3) ;fn f() = acos( + ) + bcos( + ) gks] tgk¡ a, b,  rFkk  Kkr ek=k,¡ gSa] rks – a2 + b2 + 2abcos( −) < f() < a2 + b2 + 2abcos( −) gSaA xf.kr (4) ;fn A, B, C ,d f=Hkqt ds dks.k gSa] rks sinA + sinB + sinC vkSj sinA sinB sinC dk vf/kdre eku rc gksrk gS tc A = B = C = 60° gksA (5) ;fn sin ;k cos f}?kkr ds :i esa fn;k gks rks ,d iw.kZ oxZ cukdj vf/kdre ;k U;wure ekuksa dh O;k[;k dh tk ldrh gSA SOLVED EXAMPLES Example: 16 ;fn cos 2x + 2 cos x = 1 gks] rks sin2x ( 2–cos2x) cjkcj gSA (1) 2 (2) 1 (3) 3 (4) 4- Solution: ;gk¡] cos 2x + 2 cos x = 1 ⇒ 2cos2x–1 + 2 cos x – 1 = 0 ⇒ cos2x + cosx –1 = 0 −1 + 5  −1 − 5    ;k cos x = ]   vekU;  – 1  cos x  1 vkSj  −1 − 5  < –1 2  2  2   2  5 −1  6 −2 5 cos2 x =   = = 3− 5  2  4  3 − 5   3 − 5   5 − 1  5 + 1  sin2x (2 –cos2x) =  1 −  2 −  =   = 1  2   2   2    2  1 1 Example: 17 sin 67 ° + cos 67 ° cjkcj gSA 2 2 1 1 1 1 (1) 4 +2 2 (2) 4 −2 2 (3) − 4 +2 2 (4) −4 + 2 2 2 2 2 2 1 1 1 Solution: sin 67 ° + cos 67 ° = 1 + sin135 = 1 + (cosA + sinA = 1 + sin2A dk mi;ksx djus ij) 2 2 2 1 = 4 +2 2 2    3   5   7  Example: 18  1 + cos   1 + cos   1 + cos   1 + cos  dk eku gSA  8  8   8   8  1  1 1+ 2 (1) (2) cos (3) (4) 2 8 8 2 2    3    3       Solution:  1 + cos   1 + cos   1 + cos   −    1 + cos   −    8  8    8    8       3     2   2 3  =  1 + cos   1 + cos   1 − cos   1 − cos  =  1 − cos   1 − cos   8  8  8   8   8  8  1   3  1   3  =  2 − 1 − cos   2 − 1 − cos  =  1 − cos   1 − cos  4 4  4  4 4  4   1  1  1  1 1 = 1 − 1 +   = 1 −  =   2 2  4  2 8  3 5 7 9 Example: 19 cos + cos + cos + cos + cos dk eku gSA 11 11 11 11 11 1 (1) 0 (2) 1 (3) (4) bues ls dksbZ ugh 2  f=dks.kferh; vuqikr n sin 2 cos 2 + (n − 1) Solution: ge tkurs gS cos  + cos( + ) + cos ( + 2)+…+ cos ( + n − 1 ) =  sin 2 2  2 ;gk¡  = ,  = ,n = 5 11 11  10 1   2 8  sin 5  cos  5  sin   +      11 2    cos  11 11  =     11 11   2 1   2    sin   sin   11 2     11    10          sin    sin  −   sin 1  11  1  11  1  11  1   =  =  = 2     2  sin   2  sin   2  sin 11    11   11   3 5 7 Example: 20 cos4 + cos4 + cos4 + cos4 cjkcj gS & 8 8 8 8 (1) 1/2 (2) 1/4 (3) 3/2 (4) 3/4  3 5 7  3 3  Solution: = cos4 + cos4 + cos4 + cos4 = cos4 + cos4 + cos4 + cos4 8 8 8 8 8 8 8 8 3  1   2 2   2   2 3  = 2  cos4 + cos4  =  2cos  + 2cos    8 8  2  8  8   1    2 3  2  1  2 1   1   1 2 3 =  1 + cos  +  1 + cos   =  1 +  +  1 –   = 2 + 1 = 2  4  4   2  2  2   2 2 Example: 21 ;fn A + B + C = 3 gks] rks cos 2A + cos 2B + cos 2C = 2 (1) 1 – 4 cos A cos B cos C (3) 4 sin A sin B sin C (3) 1 + 2 cos A cos B cos C (4) 1 – 4 sin A sin B sin C Solution: cos 2A + cos 2B + cos 2C = 2cos (A + B) cos (A – B) + cos 2C  3  3 = 2 cos  − C  cos (A – B) + cos 2C  A + B + C =  2  2 2 = –2 sin C cos (A – B) + 1 – 2 sin C = 1 – 2sin C [cos (A – B) + sin C)   3  = 1 – 2sin C cos (A − B) + sin  − ( A + B )   = 1 – 2 sin C [cos (A – B) – cos (A + B)]   2  = 1 – 4 sin A sin B sin C Example: 22 fdlh f=Hkqt ABC esa, sin A – cos B = cos C gks] rks dks.k B gS &     (1) (2) (3) (4) 2 3 4 6 Solution: fn;k gS ] sin A – cos B = cos C sin A = cos B + cos C xf.kr A A B+C B−C  2 sin cos = 2cos   cos   2 2  2   2  A A  −A  B−C  2 sin cos = 2 cos   cos   2 2  2   2  A+B+C= A A A B−C  2 sin cos = 2 sin cos   2 2 2  2  A B−C cos = cos ;k A = B – C 2 2 ysfdu A + B + C =  blfy, 2B =   B = /2 Example: 23 tan 9° – tan27° – tan 63° + tan 81° cjkcj gS & (1) 0 (2) 1 (3) &1 (4) 4 Solution: tan 9° + tan 81° – (tan 27° + tan 63°) (tan 9° + cot 9°) – (tan 27° + cot 27°)  sin 9º cos 9º   sin 27º cos 27º  1 1 =  + –  + = –  cos 9º sin 9º   cos 27º sin 27º  sin9º cos9º cos27º sin27º 2 2 2 2 24 2 4  5 + 1 − 5 + 1  16 = – = – = – =8 = =4 sin 18 sin 54 sin 18 sin 36º 5 −1 5 +1  ( 5 − 1)( 5 + 1)  4