www.sakshieducation.com COMPOUND ANGLES Definitions and Formulae : 1. The algebraic sum of two or more angles is called a compound a compound angle. i.e. A + B, A – B, A + B + C, A + B – C, A – B + C, B + C – A, ……. etc. are called compound angles. 2. If A and B are any two angles then i) Sin(A + B) = sin A cos B + cos A sin B. ii) Sin(A – B) = sin A cos B – cos A sin B iii) cos(A + B) = cos A cos B – sin A sin B iv) cos(A – B) = cos A cos B + sin A sin B. 3. If A, B, A + B, A – B are not odd multiples of π/2 then i) tan(A + B) = tan A + tan B . 1 − tan A tan B ii) tan(A – B) = tan A − tan B . 1 + tan A tan B 4. If A, B, A + B and A – B are not integral multiples of π, then i) cot(A + B) = cot A cot B − 1 cot B + cot A ii) cot(A – B) = cot A cot B + 1 . cot B − cot A 5. i) sin(A + B) + sin(A – B) = 2 sin A cos B ii) sin(A + B) – sin(A – B) = 2 cos A sin B iii) cos(A + B) + cos(A – B) = 2 cos A cos B iv) cos(A + B) – cos(A – B) = –2 sin A sin B v) cos(A – B) – cos(A + B) = 2sin A sin B. 6. i) sin(A + B) sin(A – B) = sin2 A – sin2 B = cos2 B – cos2 A ii) cos(A + B) cos(A – B) = cos2 – sin2 B = cos2 B – sin2 A iii) tan(A + B) tan(A – B) = iv) cot(A + B) cot(A – B) = tan 2 A − tan 2 B 1 − tan 2 A tan 2 B cot 2 Acot 2 B − 1 cot 2 B − cot 2 A www.sakshieducation.com www.sakshieducation.com cos θ + sin θ cos θ − sin θ v) tan(45o + θ) = = cot(45o – θ) = vi) tan(45o – θ) = 1 + tan θ 1 − tan θ cos θ − sin θ cos θ + sin θ = cot(45o + θ) = 1 − tan θ 1 + tan θ and tan(45o + θ) . tan(45o – θ) = 1. 7. i) sin(A + B + C) = ∑(cos A cos B cos C) – sin A sin B sin C ii) cos(A + B + C) = cos A cos B cos C – ∑(sin A sin B sin C) iii) tan(A + B + C) = ∑ (tanA) − π(tanA) 1 − ∑ (tanAtanB) VSAQ’S Simplify the following 1. cos 100° ⋅ cos 40° + sin 100° ⋅ sin 40° Sol. L.H.S. = = cos 100° ⋅ cos 40° + sin 100° ⋅ sin 40° 1 = cos (100° – 40°) = cos 60° = = R.H.S. 2 ⎛π ⎞ ⎛π ⎞ 2. tan ⎜ + θ ⎟ ⋅ tan ⎜ − θ ⎟ ⎝4 ⎠ ⎝4 ⎠ π π tan + tan θ tan − tan θ 4 4 ⋅ Sol. π π 1 − tan tan θ 1 + tan tan θ 4 4 (1 + tan θ) (1 − tan θ) = ⋅ =1 (1 − tan θ) (1 + tan θ) 3. tan 75° + cot 75° Sol. tan 75° = 2 + 3 cot 75° = 2 − 3 ∴ tan 75° + cot 75° = 2 + 3 + 2 − 3 = 4 www.sakshieducation.com www.sakshieducation.com 4. Express Sol. ( 3 cos 25° + sin 25°) as a sine of an angle. 2 ( 3 cos 25° + sin 25°) 2 3 1 cos 25° + sin 25° 2 2 = sin 60° cos 25° + cos 60° sin 25° = = sin(60° + 25°) = sin 85° 5. tan θ in terms of tan α, if sin(θ + α) = cos (θ + α). Sol. sin(θ + α) = cos (θ + α) sin(θ + α) =1 cos(θ + α) tan(θ + α) = 1 tan θ + tan α =1 1 − tan θ tan α tan θ + tan α = 1 − tan θ tan α tan θ + tan θ tan α = 1 − tan α tan θ[1 + tan α] = 1 − tan α ∴ tan θ = 1 − tan α 1 + tan α cos11° + sin11° and θ is in the third quadrant, find θ. cos11° − sin11° cos11° + sin11° Sol. tan θ = cos11° − sin11° ⎡ sin11° ⎤ cos11° ⎢1 + ⎣ cos11° ⎥⎦ = ⎡ sin11° ⎤ cos11° ⎢1 − ⎣ cos11° ⎥⎦ 1 + tan11° = 1 − tan11° tan 45° + tan11° (∵ tan 45° = 1) = 1 − tan 45° tan11° tan θ = tan(45° + 11°) 6. If tan θ = = tan 56° = tan(180° + 56°) = tan 236° θ = 236° www.sakshieducation.com www.sakshieducation.com 7. If 0° < A, B < 90°, such that cos A = Sol. cos A = 5 4 and sin B = , find the value of sin (A – B). 13 5 5 4 and sin B = 13 5 P 13 12 A Q 5 R PQ 2 = PR 2 − QR 2 = (13) 2 − 52 = 169 − 25 = 144 PQ = 12 cos A = 5 12 ,sin A = 13 13 X 5 4 Y B 3 Z YZ2 = XZ2 − XY 2 = 52 − 42 = 24 − 16 = 9 YZ = 3 4 3 sin B = , cos B = 5 5 sin(A − B) = sin A cos B − cos A sin B = 12 3 5 4 36 20 36 − 20 16 × − × = − = = 13 5 13 5 65 65 65 65 8. What is the value of tan 20° + tan 40° + Sol. tan 20° + tan 40° + 3 tan 20° tan 40°? 3 tan 20° tan 40° Consider 20° + 40° = 60° Tan(20° + 40°) = tan 60° tan 20° + tan 40° = 3 1 − tan 20° tan 40° tan 20° + tan 40° = 3(1 − tan 20° tan 40°) tan 20° + tan 40° = 3 − 3 tan 20° tan 40° tan 20° + tan 40° + 3 tan 20° tan 40° = 3 www.sakshieducation.com www.sakshieducation.com 9. Find the value of tan 56° – tan 11° – tan 56° tan 11°. Sol. We have 56° – 11° = 45° tan(56° − 11°) = tan 45° tan 56° − tan11° =1 1 + tan 56° tan11° tan 56° − tan11° = 1 + tan 56° tan11° tan 56° − tan11° − tan 56° tan11° = 1 sin(C − A) if none of sin A, sin B, sin C is zero. cos Csin A sin(C − A) sin C cos A − cos Csin A Sol. Σ =Σ sin C sin A sin Csin A sin C cos A cos C sin A =Σ − sin Csin A sin Csin A = Σ cot A − cot C = cot A − cot C + cot B − cot A + cot C − cot B = 0 10. Evaluate Σ 11. tan 72° = tan 18° + 2 tan 54° Sol. 72° – 18° = 54° Take tan on both sides tan(72° − 18°) = tan 54° tan 72° − tan18° = tan 54° 1 + tan 72° tan18° tan 72° − tan18° = tan 54° 1 + tan(90 − 18) tan18° tan 72° − tan18° = tan 54° 1 + tan18° tan18° tan 72° − tan18° = tan 54° 1+1 tan 72° − tan18° = tan 54° 2 tan 72° − tan18° = 2 tan 54° tan 72° = tan18° + 2 tan 54° 12. sin 750° cos 480° + cos 120° cos 60° = Sol. sin 750° = sin(2⋅360° + 30°) = sin 30° = −1 2 1 2 cos 480° = cos(360° + 120°) = cos 120° = 120° = cos(180° – 60°) www.sakshieducation.com www.sakshieducation.com = –cos 60° = – cos 120° = – 1 2 1 2 1 2 ∴ L.H.S. = = sin 750° cos 480° + cos 120° cos 60° 1⎛ 1⎞ ⎛ 1⎞1 = ⎜− ⎟+⎜− ⎟ 2⎝ 2⎠ ⎝ 2⎠2 cos 60° = 1 1 −2 −1 =− − = = = R.H.S. 4 4 4 2 ⎛ 4π ⎞ ⎛ 4π ⎞ 13. cos A + cos ⎜ − A ⎟ + cos ⎜ + A⎟ = 0 ⎝ 3 ⎠ ⎝ 3 ⎠ ⎛ 4π ⎞ ⎛ 4π ⎞ − A ⎟ + cos ⎜ + A⎟ Sol. Consider cos ⎜ ⎝ 3 ⎠ ⎝ 3 ⎠ = cos(240° + A) + cos(240° − A) = cos 240° cos A − sin 240° sin A + cos 240° cos A + sin 240° sin A = 2 cos 240° cos A = 2 cos(180° + 60°) cos A = −2 cos 60° cos A 1 = −2 × cos A = − cos A 2 ⎛ 2π ⎞ ⎛ 2π ⎞ 3 14. cos 2 θ + cos 2 ⎜ + θ ⎟ + cos 2 ⎜ − θ⎟ = ⎝ 3 ⎠ ⎝ 3 ⎠ 2 ⎛ 2π ⎞ ⎛ 2π ⎞ Sol. cos 2 ⎜ + θ ⎟ + cos 2 ⎜ − θ⎟ ⎝ 3 ⎠ ⎝ 3 ⎠ www.sakshieducation.com www.sakshieducation.com = cos 2 (60° + θ) + cos 2 (60° − θ) = [cos 60° cos θ − sin 60° sin θ]2 + [cos 60° cos θ + sin 60° sin θ]2 = 2(cos 2 60° cos 2 θ + sin 2 60° sin 2 θ)[∵ (a + b) 2 + (a − b) 2 = 2(a 2 + b 2 )] 2 ⎡ 1 2 ⎤ ⎛ 3⎞ ⎛ ⎞ 2 2 ⎢ = 2 ⎜ ⎟ cos θ + ⎜⎜ ⎟⎟ sin θ ⎥ ⎢⎝ 2 ⎠ ⎥ ⎝ 2 ⎠ ⎣ ⎦ 3 ⎡1 ⎤ = 2 ⎢ cos 2 θ + sin 2 θ ⎥ 4 ⎣4 ⎦ 2 = [cos 2 θ + 3sin 2 θ] 4 1 3 = cos 2 θ + sin 2 θ 2 2 1 3 ∴ L.H.S. = cos 2 θ + cos 2 θ + sin 2 θ 2 2 3 3 = cos 2 θ + sin 2 θ 2 2 3 = ⎡⎣cos 2 θ + sin 2 θ ⎤⎦ 2 3 = (∵ cos 2 θ + sin 2 θ = 1) 2 = R.H.S. 1° 1° − sin 2 22 . 2 2 1° 1° Sol. Put A = sin 2 82 and B = sin 2 22 , then 2 2 1° 1° sin 2 82 − sin 2 22 2 2 15. Evaluate sin 2 82 = sin 2 A − sin 2 B = sin(A + B) sin(A − B) = sin105° sin 60° = sin(90° + 15°) sin 60° = cos15° sin 60° = 1+ 3 3 3 + 3 ⋅ = 2 2 2 4 2 www.sakshieducation.com www.sakshieducation.com 1 1 3 +1 − sin 2 22 = 2 2 4 2 0 16. Prove that sin 2 52 0 ⎛π A⎞ ⎛π A⎞ 17. sin 2 ⎜ + ⎟ − sin 2 ⎜ − ⎟ ⎝8 2 ⎠ ⎝8 2 ⎠ ⎛π A⎞ ⎛π A⎞ Sol. sin 2 ⎜ + ⎟ − sin 2 ⎜ − ⎟ ⎝8 2 ⎠ ⎝8 2 ⎠ [∵ sin 2 A − sin 2 B = sin(A + B) sin(A − B)] ⎛π A π A⎞ ⎛π A π A⎞ = sin ⎜ + + − ⎟ sin ⎜ + − + ⎟ ⎝8 2 8 2 ⎠ ⎝8 2 8 2 ⎠ ⎛ 2π ⎞ ⎛ 2A ⎞ = sin ⎜ ⎟ sin ⎜ ⎟ ⎝ 8 ⎠ ⎝ 2 ⎠ 1 π = sin sin A = sin 45° sin A = sin A 4 2 1 1 18. cos 2 52 ° − sin 2 22 ° 2 2 1 1 Sol. cos 2 52 ° − sin 2 22 ° 2 2 [∵ cos 2 A − sin 2 B = cos(A + B) cos(A − B)] 1 ⎞ 1 ⎞ ⎛ 1 ⎛ 1 = cos ⎜ 52 ° + 22 ° ⎟ cos ⎜ 52 ° − 22 ° ⎟ 2 ⎠ 2 ⎠ ⎝ 2 ⎝ 2 = cos 75° cos 30° = 3 (cos 75°) 2 = 3 ⎛ 3 −1 ⎞ 3 − 3 ⎜ ⎟= 2 ⎜⎝ 2 2 ⎟⎠ 4 2 0 0 0 0 0 0 ⎛ ⎛ 1 1 1 1 ⎞ 1 1 ⎞ 2 19. cos 112 − sin 52 = cos ⎜112 + 52 ⎟ cos ⎜112 − 52 ⎟ 2 2 2 2 ⎠ 2 2 ⎠ ⎝ ⎝ 2 (∵ cos 2 A − sin 2 B = cos ( A + B ) cos ( A − B ) ) 1 1 = cos1650. cos 600 = cos (1800 − 150 ) = − cos150 2 2 1 ⎧⎪ 3 + 1 ⎫⎪ =− ⎨ ⎬ 2 ⎩⎪ 2 2 ⎭⎪ www.sakshieducation.com www.sakshieducation.com 20. Prove that tan 3 A tan 2 A tan A = tan 3 A − tan 2 A −tan A Solution: We know that tan 3 A = tan ( 2 A + A ) tan 2 A + tan A 1 − tan 2 A tan A tan 3 A − tan 2 A tan A tan 3 = tan 2 A + tan A tan 3 A − tan 2 − tan A = tan A tan 2 A tan 3 A tan 3 A = 21. Find the expansion of sin(A + B + C). Sol. sin(A + B + C) = sin[(A + B) − C] = sin(A + B) cos C − cos(A − B) sin C = (sin A cos B + cos A sin B) cos C − [cos(A − B) cos B − sin A sin B]sin c = sin A cos Bcos C + cos A sin Bcos C − cos A cos Bsin C + sin A sin Bsin C 22. Find the expansion of cos (A – B – C). Sol. cos (A – B – C) = cos[(A – B) – C] = cos(A − B) cos C + sin(A − B) sin C = (cos A cos B + sin A sin B) cos C + (sin A cos B − cos A sin B]sin C = cos A cos Bcos C + sin A sin Bcos C + sin A cos Bsin C − cos A sin Bsin C 23. For what values of x in the first quadrant Sol. 2 tan x is positive? 1 − tan 2 x 2 tan x > 0 ⇒ tan 2x > 0 1 − tan 2 x π ⇒ 0 < 2x < (since x in the first quadrant) 2 π ⇒0<x< 4 Therefore, π 2 tan x is positive for 0 < x < 2 4 1 − tan x www.sakshieducation.com www.sakshieducation.com SAQ’S 24. Prove that sin 4 Solution: sin 4 π 8 + sin 4 π 8 + sin 4 3π 5π 7π 3 + sin 4 + sin 4 = 8 8 8 2 3π 5π 7π + sin 4 + sin 4 8 8 8 2 4 π ⎞⎫ ⎛ π ⎞ ⎧ 2 π π ⎫ ⎧ 2 ⎛ π π ⎞⎫ ⎧ ⎛ − ⎬ + ⎨sin ⎜ + ⎟ ⎬ + ⎨sin ⎜ π − ⎟ ⎬ ⎜ sin ⎟ + ⎨sin 8⎠ ⎩ 2 8⎭ ⎩ 8 ⎠⎭ ⎝ ⎝ 2 8 ⎠⎭ ⎩ ⎝ π π π π sin 4 + cos 4 + cos 4 + sin 4 8 8 8 8 2 π π ⎪⎫ ⎪⎧⎛ 2 π ⎧ 4π 4 π ⎫ 2 π ⎞ + cos ⎬ = 2 ⎨⎜ sin + cos ⎟ − 2sin 2 cos 2 ⎬ 2 ⎨sin 8 8⎭ 8 8⎠ 8 8 ⎪⎭ ⎩ ⎪⎩⎝ π π = 2 − 4sin 2 cos 2 8 8 2 π π⎞ π 1 3 ⎛ = 2 − ⎜ 2sin cos ⎟ = 2 − sin 2 = 2 − = 8 8⎠ 4 2 2 ⎝ 4 4 5π 7π 3 + cos 4 = 8 8 8 8 2 1 26. Prove that (i) sin A sin ( 60 − A ) sin 600 + A = sin 3 A 4 1 (ii) cos A cos ( 60 + A ) cos 600 − A = cos 3 A 4 0 0 (iii) tan A tan ( 60 + A ) tan ( 60 − A ) = tan 3 A 25. Prove that cos 4 π + cos 4 3 π + cos 4 ( ) ( ) 3 16 π 2π 3π 4π 1 cos cos = (v) cos cos 9 9 9 9 16 (iv) sin 200 sin 400 sin 600sin 800 = 27. Prove that tan θ + 2 tan 2θ + 4 tan 4θ + 8cot 8θ = cot θ π −3 7 π and sin β = , where < α < π and 0 < β < , then find the values of 5 25 2 2 tan(α + β) and sin(α + β). −3 π , where < α < π Sol. cos α = 5 2 α in II quadrant 7 π sin β = , where 0 < β < 25 2 β in I Quadrant 28. If cos α = www.sakshieducation.com www.sakshieducation.com A 5 4 α B 3 C AB2 = AC2 − BC2 = 52 − 32 = 25 − 9 = 16 ∴ AB = 4 3 4 4 cos α = − ,sin α = , tan α = − 5 5 3 X 25 7 β Y 24 Z YZ2 = XZ2 − XY 2 = 252 − 7 2 = 625 − 49 = 576 YZ = 24 7 24 7 , cos β = , tan β = 25 25 24 tan α + tan β ∴ tan(α + β) = 1 − tan α tan β sin β = −4 7 −32 + 7 + 4 = 3 24 = 4 7 7 1+ × 1+ 3 24 18 −25 3 −25 18 = 24 = × =− 18 + 7 24 25 4 18 sin(α + β) = sin α cos β + cos α sin β 4 24 −3 7 = × + × 5 25 5 25 96 21 75 3 = − = = 125 125 125 5 www.sakshieducation.com www.sakshieducation.com 29. If 0 < A < B < π 24 4 and sin(A + B) = and cos(A − B) = , then find the value of 4 25 5 tan 2A. Sol. sin(A + B) = 24 25 25 24 A+B 7 24 7 4 cos(A − B) = 5 tan(A + B) = 5 3 A–B 4 3 4 Now 2A = (A + B) + (A – B) tan 2A = tan [ (A + B) + (A − B) ] tan(A − B) = = tan(A + B) + tan(A − B) 1 − tan(A + B) tan(A − B) 24 3 + 96 + 21 −117 = 7 4 = = 24 3 28 − 72 44 1− × 7 4 30. If A + B, A are acute angles such that sin ( A + B ) = 24 3 and tan A = find the value of 25 4 cos B Solution: sin ( A + B ) = 24 25 ∴cos ( A + B ) = 7 25 3 3 4 ∴ sin A = & cos A = 4 5 5 cos B = cos ( A + B − A ) = cos ( A + B ) cos A + sin ( A + B ) sin A tan A = 7 4 24 3 100 4 × + × = = 25 5 25 5 125 5 www.sakshieducation.com www.sakshieducation.com 31. If tan α – tan β = m and cot α – cot β = n, then prove that cot(α − β) = 1 1 − . m n Sol. We have tan α – tan β = m 1 1 − =m cot α cot β cot β − cot α =m cot α cot β cot α cot β 1 = cot β − cot α m cot α − cot β = n ∴ ...(1) −(cot β − cot α) = n cot β − cot α = −n 1 1 =− cot β − cot α n L.H.S. = cot(α − β) = ...(2) cot α cot β cot β − cot α = cot α cot β 1 + cot β − cot α cot β − cot α = 1 1 − (∵ from(1) & (2)) = R.H.S. m n 32. If tan(α – β) = 7 4 and tan α = , where α and β are in the firs quadrant prove that 24 3 α + β = π/2. 7 4 Sol. tan(α – β) = and tan α = 24 3 7 25 2β 34 www.sakshieducation.com www.sakshieducation.com tan(α − β) = tan α − tan β 1 + tan α tan β 4 − tan β 7 ⇒ 3 = 4 1 + tan β 24 3 4 − 3 tan β 7 3 ⇒ = 3 + 4 tan β 24 3 4 − 3 tan β 7 ⇒ = 3 + 4 tan β 24 ⇒ 24[4 − 3 tan β] = 7(3 + 4 tan β) ⇒ 96 − 72 tan β = 21 + 28 tan β ⇒ 96 − 21 = 28 tan β + 72 tan β ⇒ 100 tan β = 75 75 3 = 100 4 tan α + tan β ∴ tan(α + β) = 1 − tan α tan β ∴ tan β = 4 3 + 3 4 =α = 4 3 1− ⋅ 3 4 π tan(α + β) = tan 2 π ∴α + β = 2 33. If sin(α + β) a + b , then prove that a tan β = b tan α . = sin(α − β) a − b sin(α + β) a + b = sin(α − β) a − b By using componendo and dividendo, we get sin(α + β) + sin(α − β) a + b + a − b 2a a = = = sin(α + β) − sin(α − β) a + b − a + b 2b b Sol. Given that ⇒ sin(α + β) + sin(α − β) a = sin(α + β) − sin(α − β) b ⇒ sin α cos β + cos α sin β + sin α cos β − cos α sin β a = sin α cos β + cos α sin β − sin α cos β + cos α sin β b www.sakshieducation.com www.sakshieducation.com 2sin α cos β a = 2 cos α sin β b a ⇒ tan α cot β = b ⇒ b tan α = a tan β hence, a tan β = b tan α ⇒ 3π , then show that 4 (1 – tan A) (1 + tan B) = 2. 3π Sol. A − B = 4 A − B = 135° 34. If A − B = tan(A − B) = tan135° = tan(90° + 45)° = − cot 45° = −1 ∴ tan A − tan B = −1 1 + tan A tan B tan A − tan B = −(1 + tan A tan B) tan A − tan B = −1 − tan A tan B tan A − tan B + tan A tan B = −1 tan B − tan A − tan A tan B = 1 ...(1) L.H.S. = (1 − tan A)(1 + tan B) = 1 + (tan B − tan A − tan A tan B) = 1 + 1 (∵ from(1)) = 2 = R.H.S. π and if none of A, B, C is an odd multiple of π/2, then prove that 2 cot A + cot B + cot C = cotA cotB cotC. π Sol. A + B + C = 2 π A+B = −C 2 ⎛π ⎞ cot(A + B) = cot ⎜ − C ⎟ ⎝2 ⎠ cot A cot B − 1 = tan C cot B + cot A cot A cot B − 1 1 = cot B + cot A cot C cot C[cot A cot B − 1] = cot B + cot A 35. If A + B + C = cot A cot Bcot C − cot C cot A + cot B cot A cot Bcot C = cot A + cot B + cot C ∴ cot A + cot B + cot C = cot A cot Bcot C www.sakshieducation.com www.sakshieducation.com π and if none of A, B, C is an odd multiple of π/2, then prove that, 2 tan A tan B + tan B tan C + tan C tan A = 1. π Sol. A + B + C = 2 π A+B = −C 2 ⎛π ⎞ tan(A + B) = tan ⎜ − C ⎟ ⎝2 ⎠ 36. If A + B + C = tan A + tan B = cot C 1 − tan A tan B tan A + tan B 1 ⇒ = 1 − tan A tan B tan C ⇒ tan C[tan A + tan B] = 1 − tan A tan B ⇒ ⇒ tan C tan A + tan C tan B = 1 − tan A tan B ⇒ tan A tan B + tan B tan C + tan C tan A = 1 cos(B + C) = 2. cos Bcos C cos(B + C) Sol. L.H.S. = Σ cos Bcos C cos Bcos C − sin Bsin C =Σ cos Bcos C cos Bcos C sin Bsin C =Σ − cos Bcos C cos B cos C = Σ(1 − tan B tan C) 37. Σ = 1 − tan B tan C + 1 − tan C tan A + 1 − tan A tan B = 3 − (tan A tan B + tan B tan C + tan C tan A) = 3 − 1 (∵ from(b)) = 2 = R.H.S. 38. Prove that sin2 α + cos2 (α + β) + 2 sin α sin β cos(α + β) is independent of α. Sol. Given expression, sin2α + cos2 (α+β) + 2 sinα sinβ cos(α+β) = sin 2 α + 1 − sin 2 (α + β) + 2sin α sin β cos(α + β) = 1 + [sin 2 α − sin 2 (α + β)] + 2sin α sin β cos(α + β) = 1 + sin(α + α + β) sin(α − α − β) + 2sin α sin β cos(α + β) www.sakshieducation.com www.sakshieducation.com = 1 + sin(2α + β) sin(−β) + 2sin α sin β cos(α + β) = 1 − sin(2α + β) sin β + [2sin α cos(α + β)]sin β = 1 − sin(2α + β) sin α + [sin(α + α + β) + sin(α − α − β)]sin β = 1 − sin(2α + β) sin α + [sin(2α + β) − sin β]sin β = 1 − sin(2α + β) sin α + sin(2α + β) sin β − sin 2 β = 1 − sin 2 β = cos 2 β Thus the given expression is independent of α. π 2π 3π 7π ⋅ cot ⋅ cot ...cot = 1. 16 16 16 16 π 2π 3π 7π Sol. cot ⋅ cot ⋅ cot ...cot 16 16 16 16 7π ⎞ ⎛ 2π 6π ⎞ ⎛ 3π 5π ⎞ 4π π ⎛ = ⎜ cot ⋅ cot ⎟ ⎜ cot ⋅ cot ⎟ ⎜ cot ⋅ cot ⎟ ⋅ cot 16 ⎠ ⎝ 16 16 ⎠ ⎝ 16 16 ⎠ 16 ⎝ 16 39. Prove that cot ⎡ π π ⎛ π π ⎞ ⎤ ⎡ 2π ⎛ π 2π ⎞ ⎤ ⎡ 3π ⎛ π 3π ⎞ ⎤ = ⎢ cot ⋅ cot ⎜ − ⎟ ⎥ ⎢cot ⋅ cot ⎜ − ⎟ ⎥ ⎢cot ⋅ cot ⎜ − ⎟ ⎥ ⋅ cot 4 ⎝ 2 16 ⎠ ⎦ ⎣ 16 ⎝ 2 16 ⎠ ⎦ ⎣ 16 ⎝ 2 16 ⎠ ⎦ ⎣ 16 π π ⎞⎛ 2π 2π ⎞ ⎛ = ⎜ cot ⋅ tan ⎟ ⎜ cot ⋅ tan ⎟ 16 ⎠ ⎝ 16 16 ⎠ ⎝ 16 = 1× 1× 1× 1 = 1 3π 3π ⎞ ⎛ ⎜ cot ⋅ tan ⎟ ⋅1 16 16 ⎠ ⎝ 40. Prove that tan 70° – tan 20° = 2 tan 50°. Sol. tan 50° = tan(70° – 20°) tan 70° − tan 20° = 1 + tan 70° tan 20° ⇒ tan 70° − tan 20° = tan 50°(1 + tan 70°⋅ tan 20°) = tan 50°(1 + tan 70°⋅ tan(90° − 70°)] = tan 50°[1 + tan 70° ⋅ cot 70°] = tan 50°[1 + 1] = 2 tan 50° ∴ tan 70° − tan 20° = 2 tan 50° www.sakshieducation.com www.sakshieducation.com 41. If A + B = 45°, then prove that i) (1 + tan A) (1 + tan B) = 2 ii) (cot A – 1)(cot B – 1) = 2. Sol. i) A + B = 45° ⇒ tan(A + B) = tan 45° = 1 tan A + tan B =1 1 − tan A tan B ⇒ tan A + tan B = 1 − tan A tan B ⇒ ⇒ tan A + tan B + tan A tan B = 1...(1) Now, (1 + tan A)(1 + tan B) = 1 + tan A + tan B + tan A tan B = 2 (from(1)) ii) A + B = 45° ⇒ cot(A + B) = cot 45° = 1 cot A cot B − 1 ⇒ =1 cot B + cot A ⇒ cot A cot B − 1 = cot A + cot B ⇒ cot A cot B − cot A − cot B = 1...(2) Now, (cot A − 1)(cot B − 1) = cot A cot B − cot A − cot B + 1 = 2 (from(2)) 42. If A, B, C are the angles of a triangle and if none of them is equal to π/2, then prove that i) tan A + tan B + tan C = tan A tan B tan C ii) cotA cotB + cotB cotC + cot C cot A =1 Sol. i) Given A + B + C = π ⇒ A+B = π−C ⇒ tan(A + B) = tan(π − C) tan A + tan B = − tan C 1 − tan A tan B ⇒ tan A + tan B = − tan C(1 − tan A tan B) ⇒ ⇒ tan A + tan B = − tan C + tan A tan B tan C ⇒ tan A + tan B + tan C = tan A tan B tan C 1 etc., in (i) above, we get ii) Replacing tan A by cot A 1 1 1 1 + + = cot A cot B cot C cot A cot Bcot C ⇒ cot A cot B + cot Bcot C + cot C cot A = 1 www.sakshieducation.com www.sakshieducation.com LAQ’S 43. Let ABC be a triangle such that cot A + cot B + cot C = 3 . Then prove that ABC is an equilateral triangle. Sol. Given that A + B + C = 180° We get Σ cot A cot B = 1 Now, Σ(cot A – cot B)2 = Σcot2 A + cot2 B – 2 cot A cot B = 2 cot 2 A + 2 cot 2 B + 2 cot 2 C − 2 cot A cot B −2 cot Bcot C − 2 cot C cot A (on expanding) = 2{(cot A + cot B + cot C) 2 − 2(cot A cot B) − 2 cot Bcot C − 2 cot C cot A} −2(cot A cot B + cot Bcot C + cot C cot A) = 2(cot A + cot B + cot C) 2 − 6(cot A cot B + cot Bcot C + cot C cot A) = 2⋅3 − 6 = 0 ⇒ cot A = cot B = cot C ⇒ cot A = cot B = cot C = 3 1 = 3 3 (since cot A + cot B + cot C = 3) ⇒ A + B + C = 60° (Since each angle lies in the interval [0,180°] www.sakshieducation.com