# Trigonometry 1.TRIGONOMETRIC RATIOS

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1 Trigonometry.TRIGONOMETRIC RATIOS. If a ray OP makes an angle with the positive direction of X-axis then y x i) Sin ii) cos r r iii) tan x y (x 0) iv) cot y x (y 0) y P v) sec x r (x 0) vi) cosec y r (y 0). Relations : i) sin cosec ii) cos sec iii) tan cot iv) sin + cos v) + tan sec (sec + tan ) (sec tan ). sec + tan sec tan vi) + cot cosec (cosec + cot ) (cosec cot ) cosec + cot cosec cot vii) sec + cosec sec. cosec viii) tan sin tan. sin ; cot cos cot. cos ix) sin + cos 4 sin cos sin 4 + cos x) sin 4 + cos 4 sin cos xi) sin 6 + cos 6 sin cos xii) sin x + cosec x xiii) cos x + sec x xiv) tan x + cot x.. Values of trigonometric ratios of certain angles O r x y x angle ratio sin cos tan cot cosec sec 0 0 undefined undefined o 0 π/6 π/4 π/ π/ / / / / / / / / / / 0 undefined 0 undefined

2 4. Signs of Trigonometric ratios : If lies in I, II, III, IV quadrants then the signs of trigonometric ratios are as follows. II 90 o < < 80 o sin and cosec are (+)ve I 0 o < < 90 o all the ratios are (+)ve III 80 o < < 70 o tan and cot are (+)ve IV 70 o < < 60 o cos and sec are (+)ve Note : i) 0 o, 90 o, 80 o, 70 o. 60 o, 450 o,.. etc. are called quadrant angles. ii) With ALL SILVER TEA CUPS symbol we can remember the signs of trigonometric ratios. 5. Coterminal angles : If two angles differ by an integral multiples of 60 o then two angles are called coterminal angles. Thus 0 o, 90 o, 750 o, 0 o etc., are coterminal angles. Fn sin cos ± sin cos sin cos ± sin cos sin cos tan ± cot tan ± cot tan cosec sec ± cosec sec cosec sec ± cosec sec cosec sec cot ± tan cot ± tan cot 6. Complementary Angles : Two Angles A, B are said to complementary A + B 90 o 7. Supplementary angles : Two angles A, B are said to be supplementary A + B 80 o.

3 Convert the following into simplest form :. tan( 4π) Sol. tan(4π ) tan[7 π ] [ tan ] tan PROBLEMS VSAQ S. cosec (5π + ) Sol. cosec (5π + ) cosec ( cos ec(n π+ ) ( ) cos ec ) n 7π. Find the value of cos 7π 7π Sol. cos cos 0 π ( cos(n+ ) 0) 4. Find the value of cot(5 ) Sol. cot(5 ) cot 5 [cot ] [ cot 45 ] cot Evaluate cos 45 + cos 5 + cos 5 + cos 5 Sol. cos 45, cos5 cos(80 45 )

4 cos 45 cos 5 cos( ) cos 45 cos5 cos(60 45 ) cos 45 Given expression cos 5 sin 5 + tan 495 cot 495 Sol. cos 5 cos( ) cos 45 sin 5 sin( ) sin 45 tan 495 tan[5(90 + ) 45] cot 45 cot 495 cot[5(90 + ) 45 ] tan 45 G.E If cos t (0 < t < ) and does not lie in the first quadrant, find the value of (a) sin (b) tan. Sol. cos t lies in IV quadrant. x AC BC t x t AB a) sin t AC AB t b) tan BC t

5 8. If sin and does not lie in third quadrant, find the values of (i) cos (ii) cot Solution: sin does not lie in III Q lies in IV Quadrant cos sin 9 cos cot sin Find the value of sin 0. cos0 + cos 0. sin 00 Solution: ( ) sin 0 sin 60 0 sin cos0 cos 0 ; sin 00 sin ( ) sin 0 cos0 + cos 0. sin If cosec + cot, find cos and determine the quadrant in which lies. Sol. cosec cot (cosec + cot ) (cosec cot ) cosec cot () and cosec + cot () From () + () cosec cot cosec + cot 0 cosec + 0 cos ec 6 6 sin 6 From () () cosec cot cosec + cot

6 8 cot 8 8 cot cos cot sin Thus sin is +ve and cos is ve Then lies in II quadrant.. If sec + tan 5, find the quadrant in which lies and find the value of sin. Sol. sec + tan 5 () We know, sec tan (sec + tan )(sec tan ) sec tan sec+ tan sec tan...() 5 Adding (),() sec+ tan 5 sec tan 5 sec sec 5 6 sec 5 sec 5 tan, sin 5 sin sec and tan are positives lies in I quadrant.

7 . If cos A cos B : A does not lie in second quadrant B does not lie in third quadrant find the value of 4sin B 4tan A tan B+ sin A Solution: cos A cos B ; A does not lie in II Q A lies in III Q sin A cos B B does not lie in III Q B lies in II Q sin B tan A tan B 4 4sinB tan A tan B + sin A π π sin(π A)cos A tan A. 4 cos A π π cos ec + A sec(π+ A) cot A Sol. sin(π A) sin A π π cos A cos A sin A π tan A cot A π cos ec + A sec A sec(π+ A) sec A π π cot A cot A tan A sin A sin A cot A L.H.S. seca seca tana cos A sin A sin A sin A cos A cos A cos A 4 sin A cos A cos A sin A

8 4. π π 5π 7π 9π cot cot cot cot cot π Sol. cot cot 9 0 tan 9 π cot cot 7 0 tan 7 5π cot cot π cot cot 6 cot(90 7 ) tan 7 0 9π cot cot 8 cot(90 9 ) tan 9 0 π π 5π 7π 9π cot cot cot cot cot tan7 tan9 tan 9 tan 7 π 5π 7π sin tan sec 6 5. Simplify. 5π 7π 7π cos cos ec cos π π Sol. sin sin sin 660 sin( ) sin 60 5π tan tan050 tan( 60 0 ) tan 0 6 7π sec sec( 40 ) sec 40 sec( ) sec60 5π cos cos(5 ) cos( ) cos π csc 4 csc5 csc(60 45 ) csc 45 7π cos 6 cos 50 cos( ) cos00 cos(80 0 ) cos0

9 () L.H.S. ( ) R.H.S. tan 60 + tan 700 p 6. If tan 0 p, prove that tan 560 tan p tan 60 + tan 700 Sol. tan 560 tan 470 tan( ) + tan( ) tan( ) tan( ) tan 50 tan 40 tan 00 tan0 tan(70 0 ) tan(60 0 ) tan( ) + tan( ) cot 0 tan 0 tan 0 + cot 0 tan 0 tan 0 tan 0 + tan 0 p p p p p p + p p p + p + tan 60 + tan 700 p tan 560 tan p. 7. If α, β are complementary angles such that b sin α a, then find the value of (sin α cos β cos α sin β). Sol. α and β are complementary angles. α + β 90 β 90 α a bsin α a sin α b cos α sin a α b b a b

10 b a cos α b Since β 90 α sin β sin(90 α ) cos α cos α sin β and α 90 β sin α sin(90 β ) cosβ a cosβ sin α b a cosβ b sin α cosβ cos αsinβ a a b a b a b b b b a (b a ) a b + a a b b b b b 8. If cos + sin cos then prove that cos sin sin Solution: cos + sin cos () squaring and adding () & () Let cos sin x () ( ) ( ) ( ) cos + sin + cos sin cos + x cos x x sin cos sin sin 9. If 8 tan A 5 and 5 sin B 7 and neither A nor B is in the fourth quadrant, then show 04 that sin A cos B + cos A sin B. 45 Sol. 8 tan A 5 5 tan A 8 A lies in II quadrant 5 8 sin A,cos A 7 7 5sin B 7 7 sin B 5 B lies in III quadrant 7 4 sin B,cos B 5 5 L.H.S. : sin A cos B + cos A sin B

11 π 4π 6π 9π 0. Prove that sin + sin + sin + sin Solution: π 4π 6π 9π sin + sin + sin + sin π π π π π π sin + sin + sin + + sin π π π π π π π sin + cos + cos + sin sin + cos π π sin + cos 0 0. If 0 tan 0 λ then show that 0 0 tan 60 tan0 λ + λ 0 0 tan 60 tan 0. If sec + tan find the value of sin and determine the quadrant in which lies Solution: sec + tan sec + tan tan 5 sec + sec / sin sec lies in IV Quadrant. If A, B, C, D are angles of a cyclic quadrilateral then prove that i) sin A sin C sin D sin B ii) cos A + cos B + cos C + cos D 0 Sol. A, B, C, D are the angles of a cyclic quadrilateral. A + C 80 ; B + D 80 A 80 C, B 80 D i) sin A sin C sin D sin B sin A + sin B sin C + sin D we have A 80 C Sin A sin C () B 80 D sin B sin D () Adding () and () Sin A + sin B sin C + sin D ii) cos A cos (80 C) cos C cos A + cos C 0 () cos B cos (80 D) cos D

12 cos B + cos D 0 () Adding () and () cos A + cos B + cos C + cos D If a cos b sin c, then show that Sol. a cos b sin c () Let a sin + b cos k () Squaring and adding asin+ bcos± a + b c. 5. If sin A + 5 cos A 5, then show that 5 sin A cos A ±. Sol. sin A + 5 cos A 5 Let 5 sin A cos A k Squaring and adding ( sin A + 5 cos A) + (5 sin A cos A) 5 + k 9sin A + 5cos A + 0sin A cos A + 5sin A + 9cos A 0sin A cosa 5+ k 4sin A + 4cos A 5 + k 4(sin A + cos A) 5 + k 4() 5 + k k k k ± tan + sec + sin 6. Prove that tan sec+ cos. tan + sec Sol. We have tan sec+ (tan + sec ) (sec tan ) tan sec+ (tan + sec ) (sec+ tan )(sec tan ) tan sec+ [tan + sec ][ sec + tan ] tan sec+ tan + sec sin + sin + cos cos cos tan + sec + sin tan sec+ cos

13 7. Prove that ( + cot csc )( + tan + sec ). Sol. L.H.S. : ( + cot csc )( + tan + sec ) cos sin sin sin cos cos sin + cos cos+ sin + sin cos (sin + cos ) sin + cos + sin cos sin cos sin cos + sincos sincos R.H.S. sin cos sin cos Eliminate from the following equations. 8. x a cos ; y b sin Sol. x a cos ; y b sin x x cos cos a a y y sin sin b b / / We have cos + sin / / x y + a b 9. x a cos 4 b sin x Sol. x acos cos a x cos a 4 4 y y bsin sin b y sin b cos + sin / / x y x y + + a b a b 0. x a(sec + tan ) ; y b (sec tan ) Sol. x a(sec + tan ) ; y b (sec tan ) xy ab(sec tan ) ab () ab xy ab.

14 . Prove that cos π cot cos tan π tan sin ( π A) + A ( A) ( π + A) + A ( π A) cos A SAQ S Prove that (sin cos ) + 6(sin + cos ) + 4(sin + cos ) Sol. Consider (sin cos ) sin + cos sin cos sin cos (sin cos ) 4 [(sin cos ) ] [ sin cos ] + 4sin cos 4sin cos 4 (sin cos ) + 4sin cos 4sin cos (sin + cos ) sin + cos + sin cos + sin cos (sin + cos ) + sin cos 6 6 sin + cos (sin ) + (cos ) (sin + cos ) sin cos (sin + cos ) sin cos ( sin + cos ) 6 6 sin + cos sin cos (sin cos ) + 6(sin + cos ) + 4(sin + cos ) [+ 4sin cos 4sincos + ] 6[+ sincos + ] 4[ sin cos ] + sin cos sin cos sin cos+ 4 sin Show that cos α+ cos α sin α sec α. 4 Sol. L.H.S. : cos α+ cos α sec 4 cos α+ cos αsin α ( cos α sin α) cos α[cos α+ sin α] 4 ( sin α)[( sin α ) + sin α] ( sin α )(+ sin α) sin α R.H.S. α cos

15 ( + sin cos ) cos 4. Prove that. ( + sin + cos ) + cos Sol. Consider + sin cos ( cos ) + sin sin + sin cos sin sin cos + Again consider + sin + cos ( + cos ) + sin cos + sin cos cos cos sin + sin sin cos + L.H.S.: cos cos + sin sin sin cos R.H.S. cos cos + cos sin cos+ sin 5. If x, find the value of. + cos+ sin + sin sin Sol. x + cos+ sin sin cos x cos + sin cos 4sin cos x cos cos sin + sin x...() cos + sin

16 sin + sin cos cos+ sin + sin sin + cos + sin cos sin sin cos sin + x ( from()) sin + cos sin + cos

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