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

Leave a Reply

Your email address will not be published.

eighteen − eighteen =