text: # Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.1 Pdf
  
  ## Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.1 Pdf
  
  TN Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.1: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.1 – Trigonometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 6 Exercise 6.1 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 6.1
  
  ### TN Samacheer Kalvi 10th Maths Solutions Chapter 6 – Trigonometry
  
   Board TNSCERT Class 10th Maths Class 10 Subject Maths Chapter 6 (Exercise 6.1) Chapter Name Trigonometry
  
  ### TNSCERT Class 10th Maths Pdf | all Exercise Solution
  
  Exercise 6.1
  
  (1) Prove the following identities.
  
  (i) cotθ + tanθ = secθ cosec θ
  
  (ii) tan4θ + tan2θ = sec2θ – sec2θ
  
  Solution:
  
  No (i) Now,
  
  cotθ + tanθ
  
  = cosθ/sinθ + sinθ/cosθ [∵ tanθ = cosθ/sinθ and cotθ = cosθ/sinθ]
  
  = cos2θ + sin2θ/sinθcosθ
  
  = 1/sinθ.cosθ [∵ sin2θ + cos2θ = 1]
  
  = cosecθ.secθ [∵ 1/sinθ = cosecθ and 1/cosθ = secθ]
  
  Thus, cotθ + tanθ = cosecθ.secθ [Proved]
  
  (ii) tan4θ + tan2θ
  
  = tan2θ (tan2θ + 1)
  
  = tan2θ (sin2θ/cos2θ + 1) [∵ tanθ = sinθ/cosθ]
  
  = tan2 (sin2θ + cos2θ/cos2θ)
  
  = tan2θ × 1/cos2θ [∵ sin2θ + cos2θ = 1]
  
  = sin2θ/cos2θ × 1/cos2θ
  
  = 1-cos2θ/cos4θ [∵ sin2θ + cos2θ = 1]
  
  = 1/cos4θ – 1/cos2θ
  
  = sec4θ – sec2θ
  
  Thus, tan4θ + tan2θ = sec4θ – sec2θ [Proved]
  
  (2) Prove the following identities.
  
  (i) 1-tan2θ/cot2θ-1 = tan2θ
  
  (ii) cosθ/1+sinθ = secθ – tanθ
  
  Solution:
  
  (i) 1-tan2θ/cot2θ – 1
  
  = 1 – tan2θ/ (1/tan2θ – 1) [∵cotθ = 1/tanθ]
  
  = 1 – tan2θ/ (1-tan2θ/tan2θ)
  
  = tan2θ (1 – tan2θ)/(1 – tan2θ)
  
  = tan2θ
  
  Thus, 1 – tan2θ/cot2θ – 1 = tan2θ [Proved]
  
  (ii) Now, cosθ/1+sinθ
  
  = cosθ (1 – sinθ)/(1 + sinθ) (1 – sinθ)
  
  = cosθ – sinθ.cosθ/1 – sin2θ
  
  = cosθ – sinθ.cosθ/cos2θ [∵ sin2θ + cos2θ = 1]
  
  = cosθ/cos2θ – sinθ.cosθ/cos2θ
  
  = 1/cosθ – sinθ/cosθ
  
  = secθ – tanθ [∵ 1/cosθ = secθ and sinθ/cosθ = tanθ]
  
  Thus, cosθ/1 + sinθ = secθ – tanθ [Proved]
  
  (3) Prove the following identities.
  
  (i) √1+sinθ/1-sinθ = secθ + tanθ
  
  (ii) √1+sinθ/1-sinθ + √1-sinθ/1+sinθ = 2secθ
  
  Solution:
  
  (i) √1+sinθ/1-sinθ
  
  = √(1+sinθ) (1+sinθ)/(1-sinθ) (1+sinθ)
  
  = √(1+sinθ)2/(1-sin2θ)
  
  = √(1+sinθ)2/cos2θ
  
  = 1+sinθ/cosθ
  
  = 1/cosθ + sinθ/cosθ
  
  = secθ + tanθ
  
  Thus, √1+sinθ/1-sinθ = secθ + tanθ [Proved]
  
  (ii) √1+sinθ/1-sinθ + √1-sinθ/1+sinθ
  
  = √(1+sinθ)(1+sinθ)/(1-sinθ)(1+sinθ) + √(1-sinθ) (1-sinθ)/(1+sinθ) (1-sinθ)
  
  = √(1+sinθ)2/1-sin2θ + √(1-sinθ)2/1-sin2θ
  
  = √(1+sinθ)2/cos2θ + √(1-sinθ)2/cos2θ
  
  = 1+sinθ/cosθ + 1-sinθ/cosθ
  
  = 1 + sinθ + 1 – sinθ/cosθ = 2/cosθ = 2secθ
  
  Thus, √1+sinθ/1-sinθ + √1-sinθ/1+sinθ = 2secθ [Proved]
  
  (4) Prove the following identities.
  
  (i) sec6θ = tan6θ + 3tan2 θ sec2θ + 1
  
  (ii) (sinθ + secθ)2 + (cosθ + cosecθ)2 = 1 + (secθ + cosecθ)2
  
  Solution:
  
  (i) sec6θ
  
  = (sec2θ)3
  
  = (1 + tan2θ)[∵ sec2θ – tan2θ = 1]
  
  = 1 + 3tan2 θ + 3 (tan2θ)2 + (tan2 θ)2 [∵ (a + b)3 = a2 + 3a2b +3ab2 – b3]
  
  = 1 + 3tan2 θ + 3tan4 θ + tan6 θ
  
  = tan6 θ + 3tan2θ (1 + tan2 θ) + 1
  
  = tan6 θ + 3tan2 θ.  sec2θ + 1 [∵ sec2θ – tan2 θ = 1]
  
  = tan6θ + 3 tan2θ sec2θ + 1
  
  Thus, sec6θ = tan6θ + 3tan2θ sec2θ + 1 [Proved]
  
  (ii) (sinθ + secθ)2 + (cosθ + cosecθ)2
  
  = sin2θ + 2sinθ.secθ + sec2θ + cos2θ + 2cosθ.cosecθ + cosec2θ
  
  = (sin2θ + cos2θ) + (sec2θ + cosec2θ) + 2 (sinθ.secθ + cosθ.cosecθ)
  
  = 1 + (sec2θ + cosec2θ) + 2 (sinθ/cosθ + cosθ/sinθ) [∵ sin2θ + cos2θ = 1 secθ = 1/cosθ1cosecθ = 1/sinθ]
  
  = 1 + (sec2θ + cosec2θ) + 2 (sin2θ + cos2θ/cosθ.sinθ)
  
  = 1 + sec2θ + cosec2θ + 2 (1/cosθ. sinθ)
  
  = 1 + sec2θ + 2secθ.cosecθ + cosec2θ
  
  = 1 + (secθ + cosecθ)2 [∵ (a + b)2 = a2 + 2ab + b2]
  
  Thus, (sinθ + secθ)2 + (cosθ + cosecθ)2
  
  = 1 + (secθ + cosecθ)2 [Proved]
  
  (5) Prove the following identities.
  
  (i) sec4θ (1 – sin4θ) – 2tan2θ = 1
  
  (ii) cotθ – cosθ/cotθ + cosθ = cosecθ-1/cosecθ+1
  
  Solution:
  
  (i) sec4θ (1 – sin4θ) – 2tan2θ
  
  = sec4θ ((i)2 – (sin2θ)2) – 2 tan2θ
  
  = sec4θ {(1 – sin2θ) (1 + sin2θ)} – 2tan2θ [∵ a2 – b2 = (a+b) (a-b)]
  
  = sec4θ {cos2θ (1 + sin2θ)} – 2tan2θ
  
  = 1/cos4θ {cos2θ(1 + sin2 θ)} – 2 tan2 θ [∵secθ = 1/cosθ]
  
  = 1+sin2θ/cos2θ – 2 sin2θ/cos2θ [∵tanθ = sinθ/cosθ]
  
  = 1+sin2θ/cos2θ – 2 sin2θ/cos2θ [∵tanθ = sinθ/cosθ]
  
  = 1 + sin2θ-2 sin2θ/cos2θ
  
  = 1-sin2θ/cos2θ
  
  = cos2θ/cos2θ [∵ sin2θ + cos2θ = 1]
  
  = 1
  
  Thus, sec4θ (1 – sin4θ) – 2tan2θ = 1 [Proved]
  
  (ii) cotθ – cosθ/cotθ + cosθ
  
  = (cosθ/sinθ – cosθ)/(cosθ/sinθ + cosθ) [∵cotθ = cosθ/sinθ]
  
  = (cosθ – sinθcosθ/sinθ)/(cosθ + sinθcosθ/sinθ)
  
  = sinθ (cosθ (1 – sinθ)}/sinθ {cosθ (1 + sinθ)}
  
  = 1 – sinθ/1 + sinθ
  
  = (1 – 1/cosecθ)/(1 + 1/cosecθ) [∵sinθ = 1/cosecθ]
  
  = (cosecθ – 1/cosecθ)/cosecθ + 1/cosecθ
  
  = cosecθ (cosecθ – 1)/cosecθ (cosecθ + 1)
  
  = cosecθ – 1/cosecθ + 1
  
  Thus, cotθ – cosθ/cotθ+ cosθ = cosecθ – 1/cosec+1 [proved]
  
  (6) Prove the following identities.
  
  (i) SinA-sinB/cosA+cosB + cosA-cosB/sinA+sinB = 0
  
  (ii) sin3A+cos3A/sinA+cosA + sin3A-cos3A/sinA-cosA = 2
  
  Solution:
  
  (i) sinA–sinB/cosA+cosB + cosA– cosB/sinA+sinB
  
  = (sinA-sinB) (sinA + sinB) + (cosA-cosB) (cosA + cosB)/(cosA + cosB) (sinA + sinB)
  
  = (sin2A + cos2A) – (sin2B + cos2B)/(cosA + cosB) (sinA + sinB)
  
  = 1-1/(cosA + cosB) (sinA + sinB) = 0
  
  Thus,
  
  SinA-sinB/cosA+cosB + cosA-cosB/sinA+sinB = 0 [Proved]
  
  (ii) Sin3A+cos3A/sinA+cosA + sin3A-cos3A/sinA-cosA
  
  = (sin3A + cos3A) (sinA – cosA) + (sin3A – cos3A) (sinA + cosA)/(sinA + cosA) (sinA – cosA)
  
  = 2 (sin2A)2 – (cos2A)2}/(sin2A – cos2A)
  
  = 2 (sin2A – cos2A) (sin2A + cos2A)/(sin2A – cos2A)
  
  = 2 [∵ sin2A +cos2A = 1]
  
  Thus, sin3A+cos3A/sinA+cosA + sin3A-cos3A/sinA-cosA = 2 [Proved]
  
  (7) (i) If sinθ + cosθ = √3, then prove that tanθ + cotθ = 1.
  
  Solution:
  
  (i) Given that sinθ + cosθ = √3
  
  Now, sinθ + cosθ =√3
  
  Squaring both sides
  
  (sinθ +cosθ)2 = (√3)2
  
  sin2θ + cos2θ + 2sinθ.cosθ = 3
  
  1 + 2sinθ.cosθ = 3 [∵ sin2θ + cos2θ = 1]
  
  2sinθ.cosθ = 3 – 1 = 2
  
  sinθ.cosθ = 1 —– (i)
  
  ∴Now, tanθ + cotθ
  
  = sinθ/cosθ + cosθ/sinθ
  
  = sin2θ + cos2θ/sinθ .cosθ
  
  = 1/sinθ.cosθ [∵ sin2θ + cos2θ = 1]
  
  = 1/1 [By equation (i)
  
  = 1
  
  Thus, tanθ + cotθ = 1 [Proved]
  
  (ii) Given that √3 sinθ – cosθ = 0
  
  √3 sinθ = cosθ
  
  sinθ/cosθ = 1/√3
  
  tanθ = 1/√3
  
  tanθ = tan30°
  
  ∴θ = 30°
  
  Now, L.H.S
  
  tan 3θ
  
  = tan (3×30)°
  
  = tan 90°
  
  = undefined
  
  R.H.S
  
  3tanθ – tan3θ/1-3 tan2θ
  
  = 3tan30° – tan3 30°/1-3 tan2 30°
  
  = (3×1/√3 – 1/3)/(1-3×1/3)
  
  = (√3-1/3)/1-1
  
  = (3√3 – 1/3)/0
  
  = Undefined
  
  ∴ L.H.S = R.H.S
  
  Thus, tan3θ = 3tanθ – tan3θ/1-3 tan2θ [proved]
  
  (8) (i) If cosα/cosβ = m and cosα/sinβ = n, then prove that (m2 + n2) cos2β = n2
  
  (ii) If cotθ + tanθ = x and secθ – cosθ = y, then prove that (x2y)2/3 – (xy2)2/3 = 1
  
  Solution:
  
  Given that cosα/cosβ = m and cosα/sinβ = n
  
  Now, (m2 + n2) cos2 β
  
  = {(cosα/cosβ)2 + (cosα/sinβ)2} cos2 β
  
  = (cos2α/cos2β + cos2α/sin2β) cos2β
  
  = (cos2α sin2β + cos2α cos2β/cos2β.sin2α) cos2β
  
  = {cos2α (sin2β + cos2β/cos2β.sin2β} cos2β
  
  = cos2α/cos2β.sin2β × cos2β [∵ sin2β + cos2 β = 1]
  
  = cos2α/sin2 β
  
  = (cosα/sinβ)2
  
  = n2 [∵cosα/sinβ = n]
  
  Thus, (m2 + n2) cos2β = n[Proved]
  
  (ii) Given that cotθ + tanθ = x and secθ – cosθ = y
  
  Now, x = cotθ + tanθ
  
  x = cosθ/sinθ + sinθ/cosθ [∵cotθ = cosθ/sinθ and tanθ = sinθ/cosθ]
  
  x = cos2θ + sin2θ/sinθ.cosθ
  
  x = 1/sinθ.cosθ [∵sin2θ + cos2θ = 1]
  
  ∴ x = 1/sinθ.cosθ —– (i)
  
  Now, y = secθ – cosθ
  
  y = 1/cosθ – cosθ [∵secθ = 1/cosθ]
  
  y = 1-cos2θ/cosθ
  
  y = sin2θ/cosθ —– (ii)
  
  Now, (x2y)2/3 – (x y2)2/3
  
  = (1/sin2θ.cos2θ × sin2θ/cosθ)2/3 – (1/sinθ.cosθ × sin4θ/cos2θ)2/3 [By equation (i) and (ii)]
  
  = (1/cos3 θ)2/3 – (sin3θ/cos3θ)2/3
  
  = 1/cos2θ – sin2θ/cos2θ
  
  = 1 – sin2θ/cos2θ
  
  = cos2θ/cos2θ
  
  Thus, (x2y)2/3 – (xy2)2/3 = 1 [Proved]
  
  (9) (i) If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2 – 1) = 2p
  
  (ii) If sinθ (1 + sin2θ) = cos2θ, then prove that cos6θ – 4 cos4θ + 8 cos2θ = 4
  
  Solution:
  
  (i) Given that,
  
  sinθ + cosθ = p ——- (i)
  
  and secθ + cosecθ = q
  
  1/cosθ + 1/sinθ = q
  
  sinθ + cosθ/sinθ.cosθ =q —- (ii)
  
  Now, q (p2 – 1)
  
  = sinθ + cosθ/sinθ.cosθ (sinθ + cosθ)2 – 1} [By equation (i) and (ii)]
  
  = sinθ + cosθ/sinθ. cosθ {Sin2θ + cos2θ + 2sinθ.cosθ – 1}
  
  = sinθ + cosθ/sinθ.cosθ {1 + 2 sinθ.cosθ – 1} [∵ sin2θ + cos2θ = 1]
  
  = sinθ + cosθ/sinθ.cosθ × 2 × sinθ.cosθ
  
  = 2 sinθ + cosθ
  
  = 2 P [∵sinθ + cosθ = P]
  
  Thus, q (p2 – 1) = 2p (Proved)
  
  (ii) Given that,
  
  Sinθ (1 + sin2 θ) = cos2θ
  
  sinθ (1 + 1 – cos2θ) = cos2θ  [∵ sin2θ + cos2θ = 1]
  
  Sinθ (2 – cos2θ) = cos2θ
  
  Sinθ (2 – cos2θ)2 = cos4θ [∵ showing both sides]
  
  (1 – cos2θ) (2 – cos2θ)2 = cos4θ
  
  (1 – cos2θ) (4 – 4cos2θ + cos4 θ) = cos4θ
  
  4 – 4 cos2θ + cos4θ – 4cos2θ + 4cos4θ – cos6θ = cos4θ
  
  4 – 8 cos2θ + 4cos4θ – cos6θ = 0
  
  cos6θ – 4cos4θ + 8cos2θ = 4 [Proved]
  
  (10) If cosθ/1+sinθ = 1/a, then prove that a2-1/a2+1 = sinθ
  
  Solution:
  
  Given that cosθ/1+sinθ = 1/a
  
  Now, a2-1/a2+1
  
  = (1-1/a2)/(1+1/a2)
  
  = sin2θ + cos2θ + 2 sinθ + sin2θ – cos2θ/2 + 2sinθ
  
  = 2sinθ + 2sin2θ/2+2 sinθ
  
  = sinθ (2 + 2 sinθ)/(2 + 2 sinθ)
  
  = sinθ
  
  Thus, a2 – 1/a2 + 1 = sinθ [Proved]
  
  Here is your solution of TN Samacheer Kalvi Class 10 Maths Chapter 6 Trigonometry Exercise 6.1
  
  Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.
  
  Updated: March 2, 2022 — 3:58 pm
explanation: The document is a set of solutions for trigonometric identities from the TN Samacheer Kalvi 10th Maths textbook, specifically from Chapter 6, Exercise 6.1. Each problem presents a trigonometric identity to prove, and the solutions are provided step-by-step.
  
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distilabel_metadata.content: # Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.1 Pdf
  
  ## Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.1 Pdf
  
  TN Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.1: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.1 – Trigonometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 6 Exercise 6.1 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 6.1
  
  ### TN Samacheer Kalvi 10th Maths Solutions Chapter 6 – Trigonometry
  
   Board TNSCERT Class 10th Maths Class 10 Subject Maths Chapter 6 (Exercise 6.1) Chapter Name Trigonometry
  
  ### TNSCERT Class 10th Maths Pdf | all Exercise Solution
  
  Exercise 6.1
  
  (1) Prove the following identities.
  
  (i) cotθ + tanθ = secθ cosec θ
  
  (ii) tan4θ + tan2θ = sec2θ – sec2θ
  
  Solution:
  
  No (i) Now,
  
  cotθ + tanθ
  
  = cosθ/sinθ + sinθ/cosθ [∵ tanθ = cosθ/sinθ and cotθ = cosθ/sinθ]
  
  = cos2θ + sin2θ/sinθcosθ
  
  = 1/sinθ.cosθ [∵ sin2θ + cos2θ = 1]
  
  = cosecθ.secθ [∵ 1/sinθ = cosecθ and 1/cosθ = secθ]
  
  Thus, cotθ + tanθ = cosecθ.secθ [Proved]
  
  (ii) tan4θ + tan2θ
  
  = tan2θ (tan2θ + 1)
  
  = tan2θ (sin2θ/cos2θ + 1) [∵ tanθ = sinθ/cosθ]
  
  = tan2 (sin2θ + cos2θ/cos2θ)
  
  = tan2θ × 1/cos2θ [∵ sin2θ + cos2θ = 1]
  
  = sin2θ/cos2θ × 1/cos2θ
  
  = 1-cos2θ/cos4θ [∵ sin2θ + cos2θ = 1]
  
  = 1/cos4θ – 1/cos2θ
  
  = sec4θ – sec2θ
  
  Thus, tan4θ + tan2θ = sec4θ – sec2θ [Proved]
  
  (2) Prove the following identities.
  
  (i) 1-tan2θ/cot2θ-1 = tan2θ
  
  (ii) cosθ/1+sinθ = secθ – tanθ
  
  Solution:
  
  (i) 1-tan2θ/cot2θ – 1
  
  = 1 – tan2θ/ (1/tan2θ – 1) [∵cotθ = 1/tanθ]
  
  = 1 – tan2θ/ (1-tan2θ/tan2θ)
  
  = tan2θ (1 – tan2θ)/(1 – tan2θ)
  
  = tan2θ
  
  Thus, 1 – tan2θ/cot2θ – 1 = tan2θ [Proved]
  
  (ii) Now, cosθ/1+sinθ
  
  = cosθ (1 – sinθ)/(1 + sinθ) (1 – sinθ)
  
  = cosθ – sinθ.cosθ/1 – sin2θ
  
  = cosθ – sinθ.cosθ/cos2θ [∵ sin2θ + cos2θ = 1]
  
  = cosθ/cos2θ – sinθ.cosθ/cos2θ
  
  = 1/cosθ – sinθ/cosθ
  
  = secθ – tanθ [∵ 1/cosθ = secθ and sinθ/cosθ = tanθ]
  
  Thus, cosθ/1 + sinθ = secθ – tanθ [Proved]
  
  (3) Prove the following identities.
  
  (i) √1+sinθ/1-sinθ = secθ + tanθ
  
  (ii) √1+sinθ/1-sinθ + √1-sinθ/1+sinθ = 2secθ
  
  Solution:
  
  (i) √1+sinθ/1-sinθ
  
  = √(1+sinθ) (1+sinθ)/(1-sinθ) (1+sinθ)
  
  = √(1+sinθ)2/(1-sin2θ)
  
  = √(1+sinθ)2/cos2θ
  
  = 1+sinθ/cosθ
  
  = 1/cosθ + sinθ/cosθ
  
  = secθ + tanθ
  
  Thus, √1+sinθ/1-sinθ = secθ + tanθ [Proved]
  
  (ii) √1+sinθ/1-sinθ + √1-sinθ/1+sinθ
  
  = √(1+sinθ)(1+sinθ)/(1-sinθ)(1+sinθ) + √(1-sinθ) (1-sinθ)/(1+sinθ) (1-sinθ)
  
  = √(1+sinθ)2/1-sin2θ + √(1-sinθ)2/1-sin2θ
  
  = √(1+sinθ)2/cos2θ + √(1-sinθ)2/cos2θ
  
  = 1+sinθ/cosθ + 1-sinθ/cosθ
  
  = 1 + sinθ + 1 – sinθ/cosθ = 2/cosθ = 2secθ
  
  Thus, √1+sinθ/1-sinθ + √1-sinθ/1+sinθ = 2secθ [Proved]
  
  (4) Prove the following identities.
  
  (i) sec6θ = tan6θ + 3tan2 θ sec2θ + 1
  
  (ii) (sinθ + secθ)2 + (cosθ + cosecθ)2 = 1 + (secθ + cosecθ)2
  
  Solution:
  
  (i) sec6θ
  
  = (sec2θ)3
  
  = (1 + tan2θ)[∵ sec2θ – tan2θ = 1]
  
  = 1 + 3tan2 θ + 3 (tan2θ)2 + (tan2 θ)2 [∵ (a + b)3 = a2 + 3a2b +3ab2 – b3]
  
  = 1 + 3tan2 θ + 3tan4 θ + tan6 θ
  
  = tan6 θ + 3tan2θ (1 + tan2 θ) + 1
  
  = tan6 θ + 3tan2 θ.  sec2θ + 1 [∵ sec2θ – tan2 θ = 1]
  
  = tan6θ + 3 tan2θ sec2θ + 1
  
  Thus, sec6θ = tan6θ + 3tan2θ sec2θ + 1 [Proved]
  
  (ii) (sinθ + secθ)2 + (cosθ + cosecθ)2
  
  = sin2θ + 2sinθ.secθ + sec2θ + cos2θ + 2cosθ.cosecθ + cosec2θ
  
  = (sin2θ + cos2θ) + (sec2θ + cosec2θ) + 2 (sinθ.secθ + cosθ.cosecθ)
  
  = 1 + (sec2θ + cosec2θ) + 2 (sinθ/cosθ + cosθ/sinθ) [∵ sin2θ + cos2θ = 1 secθ = 1/cosθ1cosecθ = 1/sinθ]
  
  = 1 + (sec2θ + cosec2θ) + 2 (sin2θ + cos2θ/cosθ.sinθ)
  
  = 1 + sec2θ + cosec2θ + 2 (1/cosθ. sinθ)
  
  = 1 + sec2θ + 2secθ.cosecθ + cosec2θ
  
  = 1 + (secθ + cosecθ)2 [∵ (a + b)2 = a2 + 2ab + b2]
  
  Thus, (sinθ + secθ)2 + (cosθ + cosecθ)2
  
  = 1 + (secθ + cosecθ)2 [Proved]
  
  (5) Prove the following identities.
  
  (i) sec4θ (1 – sin4θ) – 2tan2θ = 1
  
  (ii) cotθ – cosθ/cotθ + cosθ = cosecθ-1/cosecθ+1
  
  Solution:
  
  (i) sec4θ (1 – sin4θ) – 2tan2θ
  
  = sec4θ ((i)2 – (sin2θ)2) – 2 tan2θ
  
  = sec4θ {(1 – sin2θ) (1 + sin2θ)} – 2tan2θ [∵ a2 – b2 = (a+b) (a-b)]
  
  = sec4θ {cos2θ (1 + sin2θ)} – 2tan2θ
  
  = 1/cos4θ {cos2θ(1 + sin2 θ)} – 2 tan2 θ [∵secθ = 1/cosθ]
  
  = 1+sin2θ/cos2θ – 2 sin2θ/cos2θ [∵tanθ = sinθ/cosθ]
  
  = 1+sin2θ/cos2θ – 2 sin2θ/cos2θ [∵tanθ = sinθ/cosθ]
  
  = 1 + sin2θ-2 sin2θ/cos2θ
  
  = 1-sin2θ/cos2θ
  
  = cos2θ/cos2θ [∵ sin2θ + cos2θ = 1]
  
  = 1
  
  Thus, sec4θ (1 – sin4θ) – 2tan2θ = 1 [Proved]
  
  (ii) cotθ – cosθ/cotθ + cosθ
  
  = (cosθ/sinθ – cosθ)/(cosθ/sinθ + cosθ) [∵cotθ = cosθ/sinθ]
  
  = (cosθ – sinθcosθ/sinθ)/(cosθ + sinθcosθ/sinθ)
  
  = sinθ (cosθ (1 – sinθ)}/sinθ {cosθ (1 + sinθ)}
  
  = 1 – sinθ/1 + sinθ
  
  = (1 – 1/cosecθ)/(1 + 1/cosecθ) [∵sinθ = 1/cosecθ]
  
  = (cosecθ – 1/cosecθ)/cosecθ + 1/cosecθ
  
  = cosecθ (cosecθ – 1)/cosecθ (cosecθ + 1)
  
  = cosecθ – 1/cosecθ + 1
  
  Thus, cotθ – cosθ/cotθ+ cosθ = cosecθ – 1/cosec+1 [proved]
  
  (6) Prove the following identities.
  
  (i) SinA-sinB/cosA+cosB + cosA-cosB/sinA+sinB = 0
  
  (ii) sin3A+cos3A/sinA+cosA + sin3A-cos3A/sinA-cosA = 2
  
  Solution:
  
  (i) sinA–sinB/cosA+cosB + cosA– cosB/sinA+sinB
  
  = (sinA-sinB) (sinA + sinB) + (cosA-cosB) (cosA + cosB)/(cosA + cosB) (sinA + sinB)
  
  = (sin2A + cos2A) – (sin2B + cos2B)/(cosA + cosB) (sinA + sinB)
  
  = 1-1/(cosA + cosB) (sinA + sinB) = 0
  
  Thus,
  
  SinA-sinB/cosA+cosB + cosA-cosB/sinA+sinB = 0 [Proved]
  
  (ii) Sin3A+cos3A/sinA+cosA + sin3A-cos3A/sinA-cosA
  
  = (sin3A + cos3A) (sinA – cosA) + (sin3A – cos3A) (sinA + cosA)/(sinA + cosA) (sinA – cosA)
  
  = 2 (sin2A)2 – (cos2A)2}/(sin2A – cos2A)
  
  = 2 (sin2A – cos2A) (sin2A + cos2A)/(sin2A – cos2A)
  
  = 2 [∵ sin2A +cos2A = 1]
  
  Thus, sin3A+cos3A/sinA+cosA + sin3A-cos3A/sinA-cosA = 2 [Proved]
  
  (7) (i) If sinθ + cosθ = √3, then prove that tanθ + cotθ = 1.
  
  Solution:
  
  (i) Given that sinθ + cosθ = √3
  
  Now, sinθ + cosθ =√3
  
  Squaring both sides
  
  (sinθ +cosθ)2 = (√3)2
  
  sin2θ + cos2θ + 2sinθ.cosθ = 3
  
  1 + 2sinθ.cosθ = 3 [∵ sin2θ + cos2θ = 1]
  
  2sinθ.cosθ = 3 – 1 = 2
  
  sinθ.cosθ = 1 —– (i)
  
  ∴Now, tanθ + cotθ
  
  = sinθ/cosθ + cosθ/sinθ
  
  = sin2θ + cos2θ/sinθ .cosθ
  
  = 1/sinθ.cosθ [∵ sin2θ + cos2θ = 1]
  
  = 1/1 [By equation (i)
  
  = 1
  
  Thus, tanθ + cotθ = 1 [Proved]
  
  (ii) Given that √3 sinθ – cosθ = 0
  
  √3 sinθ = cosθ
  
  sinθ/cosθ = 1/√3
  
  tanθ = 1/√3
  
  tanθ = tan30°
  
  ∴θ = 30°
  
  Now, L.H.S
  
  tan 3θ
  
  = tan (3×30)°
  
  = tan 90°
  
  = undefined
  
  R.H.S
  
  3tanθ – tan3θ/1-3 tan2θ
  
  = 3tan30° – tan3 30°/1-3 tan2 30°
  
  = (3×1/√3 – 1/3)/(1-3×1/3)
  
  = (√3-1/3)/1-1
  
  = (3√3 – 1/3)/0
  
  = Undefined
  
  ∴ L.H.S = R.H.S
  
  Thus, tan3θ = 3tanθ – tan3θ/1-3 tan2θ [proved]
  
  (8) (i) If cosα/cosβ = m and cosα/sinβ = n, then prove that (m2 + n2) cos2β = n2
  
  (ii) If cotθ + tanθ = x and secθ – cosθ = y, then prove that (x2y)2/3 – (xy2)2/3 = 1
  
  Solution:
  
  Given that cosα/cosβ = m and cosα/sinβ = n
  
  Now, (m2 + n2) cos2 β
  
  = {(cosα/cosβ)2 + (cosα/sinβ)2} cos2 β
  
  = (cos2α/cos2β + cos2α/sin2β) cos2β
  
  = (cos2α sin2β + cos2α cos2β/cos2β.sin2α) cos2β
  
  = {cos2α (sin2β + cos2β/cos2β.sin2β} cos2β
  
  = cos2α/cos2β.sin2β × cos2β [∵ sin2β + cos2 β = 1]
  
  = cos2α/sin2 β
  
  = (cosα/sinβ)2
  
  = n2 [∵cosα/sinβ = n]
  
  Thus, (m2 + n2) cos2β = n[Proved]
  
  (ii) Given that cotθ + tanθ = x and secθ – cosθ = y
  
  Now, x = cotθ + tanθ
  
  x = cosθ/sinθ + sinθ/cosθ [∵cotθ = cosθ/sinθ and tanθ = sinθ/cosθ]
  
  x = cos2θ + sin2θ/sinθ.cosθ
  
  x = 1/sinθ.cosθ [∵sin2θ + cos2θ = 1]
  
  ∴ x = 1/sinθ.cosθ —– (i)
  
  Now, y = secθ – cosθ
  
  y = 1/cosθ – cosθ [∵secθ = 1/cosθ]
  
  y = 1-cos2θ/cosθ
  
  y = sin2θ/cosθ —– (ii)
  
  Now, (x2y)2/3 – (x y2)2/3
  
  = (1/sin2θ.cos2θ × sin2θ/cosθ)2/3 – (1/sinθ.cosθ × sin4θ/cos2θ)2/3 [By equation (i) and (ii)]
  
  = (1/cos3 θ)2/3 – (sin3θ/cos3θ)2/3
  
  = 1/cos2θ – sin2θ/cos2θ
  
  = 1 – sin2θ/cos2θ
  
  = cos2θ/cos2θ
  
  Thus, (x2y)2/3 – (xy2)2/3 = 1 [Proved]
  
  (9) (i) If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2 – 1) = 2p
  
  (ii) If sinθ (1 + sin2θ) = cos2θ, then prove that cos6θ – 4 cos4θ + 8 cos2θ = 4
  
  Solution:
  
  (i) Given that,
  
  sinθ + cosθ = p ——- (i)
  
  and secθ + cosecθ = q
  
  1/cosθ + 1/sinθ = q
  
  sinθ + cosθ/sinθ.cosθ =q —- (ii)
  
  Now, q (p2 – 1)
  
  = sinθ + cosθ/sinθ.cosθ (sinθ + cosθ)2 – 1} [By equation (i) and (ii)]
  
  = sinθ + cosθ/sinθ. cosθ {Sin2θ + cos2θ + 2sinθ.cosθ – 1}
  
  = sinθ + cosθ/sinθ.cosθ {1 + 2 sinθ.cosθ – 1} [∵ sin2θ + cos2θ = 1]
  
  = sinθ + cosθ/sinθ.cosθ × 2 × sinθ.cosθ
  
  = 2 sinθ + cosθ
  
  = 2 P [∵sinθ + cosθ = P]
  
  Thus, q (p2 – 1) = 2p (Proved)
  
  (ii) Given that,
  
  Sinθ (1 + sin2 θ) = cos2θ
  
  sinθ (1 + 1 – cos2θ) = cos2θ  [∵ sin2θ + cos2θ = 1]
  
  Sinθ (2 – cos2θ) = cos2θ
  
  Sinθ (2 – cos2θ)2 = cos4θ [∵ showing both sides]
  
  (1 – cos2θ) (2 – cos2θ)2 = cos4θ
  
  (1 – cos2θ) (4 – 4cos2θ + cos4 θ) = cos4θ
  
  4 – 4 cos2θ + cos4θ – 4cos2θ + 4cos4θ – cos6θ = cos4θ
  
  4 – 8 cos2θ + 4cos4θ – cos6θ = 0
  
  cos6θ – 4cos4θ + 8cos2θ = 4 [Proved]
  
  (10) If cosθ/1+sinθ = 1/a, then prove that a2-1/a2+1 = sinθ
  
  Solution:
  
  Given that cosθ/1+sinθ = 1/a
  
  Now, a2-1/a2+1
  
  = (1-1/a2)/(1+1/a2)
  
  = sin2θ + cos2θ + 2 sinθ + sin2θ – cos2θ/2 + 2sinθ
  
  = 2sinθ + 2sin2θ/2+2 sinθ
  
  = sinθ (2 + 2 sinθ)/(2 + 2 sinθ)
  
  = sinθ
  
  Thus, a2 – 1/a2 + 1 = sinθ [Proved]
  
  Here is your solution of TN Samacheer Kalvi Class 10 Maths Chapter 6 Trigonometry Exercise 6.1
  
  Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.
  
  Updated: March 2, 2022 — 3:58 pm
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a set of solutions for trigonometric identities from the TN Samacheer Kalvi 10th Maths textbook, specifically from Chapter 6, Exercise 6.1. Each problem presents a trigonometric identity to prove, and the solutions are provided step-by-step.
  
  The document contains multiple questions and their corresponding answers.
  
  Label: multiple-question-and-answer