Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.1 Pdf
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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
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