Class 11 RD Sharma Solutions – Chapter 7 Trigonometric Ratios of Compound Angles – Exercise 7.1 | Set 2

Question 17. Prove that:

(i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x 

(ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1

(iii) tan 36° + tan 9° + tan 36° tan 9° = 1 

(iv) tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

Solution:

(i) Prove: tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

Proof:

Let’s solve LHS 

= tan 8x – tan 6x – tan 2x

= tan 8x

= tan(6x + 2x)

As we know that

tan(A + B) = (tanA + tanB) / (1 – tanA tanB)

So,

= tan 8x (tan 6x + tan 2x)/(1 tan 6x tan 2x)

Now, by cross-multiplying we get,

= tan 8x (1 – tan 6x tan 2x) = tan 6x + tan 2x

= tan 8x – tan 8x tan 6x tan2x = tan 6x + tan 2x

After rearranging we get,

= tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

= RHS

LHS = RHS 

Hence proved.

(ii) Prove: tan π/12 + tan π/6 + tan π/12 tan π/6 = 1

Proof:

As we know that

π/12 15° and π/6 = 30° 

 So, we have 15° + 30° = 45°

tan (15° +30°) = tan 45° 

Since, tan (A + B)= (tan A+ tan B) / (1 – tanA tanB)

So,

(tan 15°+tan 30°)/(1-tan 15° tan 30°) = 1 

tan 15° tan 30° = 1 – tan 15° tan 30° 

After rearranging we get, 

tan15° + tan30° + tan 15° tan30° = 1

Hence proved.

(iii) Prove: tan 36° + tan 9° + tan 36° tan 9° = 1

Proof:

As we know that

36° + 9° = 45°

tan (36° + 9°) = tan 45° 

Since, tan (A + B) = (tan A + tan B)/(1 – tanA tanB)

So, 

(tan 36° + tan 9°)/(1 – tan 36° tan 9°) = 1 

tan 36° + tan 9° = 1 – tan 36° tan 9° 

After rearranging we get, 

tan 36° + tan 9° + tan 36° tan 9° = 1 = RHS

LHS = RHS

Hence proved.

(iv) Prove: tan 13x-tan 9x-tan 4x = tan 13x tan 9x tan 4x 

Proof:

Let solve LHS, 

= tan 13x – tan 9x -tan 4x

⇒ tan 13x = tan (9x + 4x) 

We know,

tan(A + B) = (tanA + tanB)/(1 – tanA tanB)

So,

tan 13x = (tan 9x + tan 4x)/(1 – tan 9x tan 4x) 

Now by cross-multiplying we get,

tan 13x (1-tan 9x tan 4x) = tan 9x + tan 4x 

tan 13x – tan 13x tan 9x tan 4x = tan 9x + tan 4x

After rearranging we get,

tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x = RHS

LHS = RHS

Hence proved.

Question 18. Proved that

Solution:

Prove: 

Proof:

Le’s solve RHS,

= tan3θ tanθ

= tan(2θ + θ) x tan(2θ – θ)

= 

= 

= LHS

LHS = RHS

Hence proved

Question 19. If  , show that tanx/tany = a/b

Solution:

Given that 

⇒ 

⇒ 

Now by using componendo and Dividendo, we get

⇒ 

⇒ tanx/tany = a/b

Hence Proved.

Question 20. If tanA = x tanB, prove that 

Solution:

Given that

tanA = x tanB

 sinA/cosA = x sinB/cosB

⇒ sinAcosB = x cosA sinB

Now, 

= 

= 

= (x – 1)(x + 1)

Hence Proved.

Question 21. If tan(A + B) = x and tan(A – B) = y, find the values of tan2A and tan2B.

Solution:

Given that

tan(A + B) = x and tan(A – B) = y

As we know that tan2A = tan[(A+B) + (A-B)]

= 

= (x + y) / (1 – xy)

Since, tan2B = tan[(A + B) – (A – B)]

So, 

= 

= (x – y) / (1 + xy) 

Question 22. If cosA + sinB = m and sinA + sinB = n, prove that 2sin(A + B) = m² + n² – 2

Solution:

Given that

cosA + sinB = m and sinA + cosB = n

Prove: 2sin(A + B) = m² + n² – 2

Proof: 

Let’s solve RHS, m² + n² – 2

= (cosA + sinB)² + (sinA + cosB)² – 2

= cos²A + sin²B + 2cosA sinB + sin²A + cos²B + 2 sinA cosB – 2

= (sin²A + cos²A) + (sin²B + cos²B) + 2 cosA sinB + 2 sinA cosB – 2

= 1 + 1 + 2 cosA sinB + 2 sinA cosB – 2

= 2 + 2(sinA cosB + cosA sinB) – 2

= 2(sinA cosB + cosA sinB)

= 2 sin(A + B)

LHS = RHS

Hence Proved.

Question 23. If tanA + tanB = a and cotA + cotB = b, prove that cot(A + B) = 1/a – 1/b.

Solution:

Given that

tanA + tanB = a and cotA + cotB = b

Prove: cot(A + B) = 1/a – 1/b.

Proof:

Lets solve cotA + cotB = b

⇒ 1/tanA + 1/tanB = b

⇒ (tanA + tanB)/(tanA tanB) = b

⇒ a/(tanA tanB) = b

⇒ a/b = tanA tanB

Now lwts solve LHS = cot(A + B) = 1/ tan(A + B)

= 1 / (tanA + tanB)/(1 – tanA tanB)

= (1 – tanA tanB)(tanA + tanB)

= (1 –  a/b) / a

= (b-a)/ab

= b/ab – a/ab

= 1/a – 1/b

Hence proved.

Question 24. If θ lies in the first quadrant and cosθ = 8/17, then prove that:

cos(π/6 + θ) + cos(π/4 – θ) + cos(2π/3 – θ) = {(√3 – 1)/2 + 1/√2}23/17.

Solution:

Given,

0 < x < π/2

Now, sinx = 

Let’s solve LHS = cos(π/6 + x) + cos(π/4 – x) + cos(2π/3 – x)

= cos 30° cosx – sin 30° sinx + cos 45° cosx + sin 45° sinx + 
   cos 120° cosx + sin 120° sinx 

= cosx (cos 30° + cos 45° + cos 120°) + sinx (- sin 30° + sin 45° + sin 120°)

= (8/17)(√3/2 + 1/√2 – 1/2) + (15/17)(-1/2 + 1/√2 + √3/2)

= (8/17)((√3-1)/2 + 1/√2) + (15/17)((√3 – 1)/2 + 1/√2)

= (23/17)((√3-1)/2 + 1/√2)

= RHS

LHS = RHS

Hence proved

Question 25. tanx + tan(x + π/3) + tan(x + 2π/3) = 3, then prove that (3tanx – tan³x)/(1 – 3tan²x) = 1

Solution:

Given,

tanx + tan(x + π/3) + tan(x + 2π/3) = 3

Prove: (3tanx – tan³x)/(1 – 3tan²x) = 1

Proof:

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Hence proved.

Question 26. If sin(α + β) = 1 and sin(α – β) = 1/2, where 0 ≤ α, β ≤ π/2, then find the values of tan(α + 2β) and tan(2α + β)

 Solution:

Given,

sin(α + β) = 1 and sin(α – β) = 1/2

Find the values of tan(α + 2β) and tan(2α + β)

So, 

⇒ α + β = 90°   …..(i)

and α – β = 30°  …..(ii)

Now by adding eq (i) and eq (ii) we get,

⇒ 2α = 120°

⇒ α = 60°

And on subtracting eq (ii) from eq (i), we get,

⇒ 2β = 60°

⇒ β = 30°

So, 

tan(α + 2β) = tan(60° + 2 × 30°) = tan120° = -√3

tan(2α + β) = tan(2 × 60° + 30°) = tan150° = -1/√3

Question 27. If α, β are two different values of x lying between 0 and 2π, which satisfy the equation 6cosx + 8sinx = 9, find the value of sin(α + β).

Solution:

Given,

6 cosx + 8 sinx = 9

⇒ 6 cosx = 9 – 8 sinx

⇒ 36 cos²x = (9 – 8 sinx)²

⇒ 36(1 – sin²x) = 81 + 64sin²x – 144 sinx

⇒ 100 sin²x – 144 sinx + 45 = 0

Now, let us considered α and β are the roots of the given equation,

So, cosα and cosβ are the roots of the above equation.

⇒ sinα sinβ = 45/100

Again,

6cosx + 8sinx = 9

⇒ 8sinx = 9 – 6 cosx

⇒ 64 sin²x = (9 – 6 cosx)²

⇒ 64(1 – cos²x) = 81 + 36 cos²x – 108 cosx

⇒ 100 cos²x – 108 cosx + 17 = 0

Now, let us considered α and β are the roots of the given equation, 

So, sinα and sinβ are the roots of the above equation.

so, cosα cosβ = 17/100

Hence, cos(α + β) = cosα cosβ – sinα sinβ

= 17/100 – 45/100

= -28/100

= -7/25

sin(α + β) = √(1 – cos²(α + β))

= √(1 – (-7/25)²)

= √(576/625

= 24/25

Question 28 (i), If sinα + sinβ = a and cosα + cosβ = b, show that sin(α + β) = 2ab/(a² + b²)

Solution:

Given that, sinα + sinβ = a and cosα + cosβ = b

Show : sin(α + β) = 2ab/(a² + b²)

So, now solve b² + a² = (cosα + coβ)² + (sinα + sinβ)²

= (cos²α + sin²α) + (sin²β + cos²β) + 2(cosα cosβ + sinα sinβ)

= 1 + 1 + 2 cos(α – β)

= 2 + 2 cos(α – β) ……..(i)

and,

b² – a² = (cosα + coβ)² – (sinα + sinβ)²

= cos²α + cos²β – sin²α – sin²β + 2(cosα cosβ – sinα sinβ)

= (cos²α – sin²α) + (cos²β – sin²β) + 2 cos(α + β)

= 2cos(α + β)cos(α – β) + 2cos(α + β)

= cos(α + β){2cos(α – β) + 2}

= cos(α + β)(b² + a²)    …….(ii)

⇒ (b² – a²)/(b² + a²) = cos(α + β)

⇒ sin(α + β) = √(1 – cos²(α + β))

= 

= 2ab/(a² + b²)

Question 28 (ii). If sinα + sinβ = a and cosα + cosβ= b, show that cos(α + β) = (b² – a²)/(b² + a²)

Solution:

Given that, sinα + sinβ = a and cosα + cosβ= b

Show: cos(α + β) = (b² – a²)/(b² + a²)

So, now solve b² + a² = (cosα + coβ)² + (sinα + sinβ)²

= (cos²α + sin²α) + (sin²β + cos²β) + 2(cosα cosβ + sinα sinβ)

= 1 + 1 + 2 cos(α – β)

= 2 + 2 cos(α – β)  ……(i)

and,

b² – a² = (cosα + coβ)² – (sinα + sinβ)²

= cos²α + cos²β – sin²α – sin²β + 2(cosα cosβ – sinα sinβ)

= (cos²α – sin²α) + (cos²β – sin²β) – 2 cos(α + β)

= 2cos(α + β) cos(α – β) + 2cos(α – β)

= cos(α + β) {2cos(α – β) + 2}    ……..(ii)

Now from (i) and (ii), we have  

⇒ b² – a² = cos(α + β)(a² + b²)

⇒ (b² – a²)/(b² + a²) = cos(α + β)

Question 29 (i). Proved that 

Solution:

Let’s solve RHS

= 

= 

= 

= 

= 

= 

= LHS

LHS = RHS

Hence proved.

Question 29 (ii). Proved that 

Solution:

Let’s solve RHS

= 

= 

= 

= 

= 

= 

= RHS

LHS = RHS

Hence Proved.

Question 29 (iii). Proved that 

Solution:

Let’s solve RHS

= 

= 

= 

= 

= 

= 

= LHS

LHS = RHS

Hence proved  

Question 30. If sinα sinβ – cosα cosβ + 1 = 0, proved that 1 + cotα tanβ = 0

Solution:

Given,

sinα sinβ – cosα cosβ + 1 = 0

⇒ -(cosα cosβ – sinα sinβ) + 1 = 0

⇒ -cos(α + β) + 1 = 0

⇒ cos(α + β) = 1

Therefore, sin(α + β) = 0   ……(i) 

Let’s solve LHS 

= 1 + cotα tanβ = 1 + (cosα sinβ)/(sinα cosβ)

= (sinα cosβ + cosα sinβ)/ (sinα cosβ)

= sin(α + β)/ (sinα cosβ)

Now from eq(i), we get

= 0 

LHS = RHS

Hence Proved.

Question 31. tanα = x + 1 and tanβ = x – 1, show that 2cot(α – β) = x²

Solution:

We have, 

tanα = x + 1 and tanβ = x – 1

As we know that tan(α – β) = (tanα – tanβ) / (1 + tanα tanβ)

= [(x + 1) – (x – 1)] / [1 + (x + 1)(x – 1)]

= (x + 1 – x + 1) / (1 + x² – 1)

= 2/ (1 + x² – 1)

= 2/x²

cot(α – β) = x²/2

2cot(α – β) = x²

LHS = RHS

Hence Proved.

Question 32. If angle θ is divided into two parts such that the tangents of one part is λ part times the tangent of the other, and ϕ is their difference then show that sinθ = (λ + 1)/(λ – 1) sinϕ.

Solution:

Let us considered α and β be the two parts of the angle be θ. 

Then, θ = α + β and ϕ = α – β

According to question, we get

tanα = λ tanβ           

⇒ tanα / tanβ = λ/1

Now, applying componendo and dividendo, we get

⇒ (tanα + tanβ) / (tanα – tanβ) = (λ+1) / (λ-1)          

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Hence proved.

Question 33. If tanθ = (sinα – cosα)/(sinα + cosα), then show that sinα + cosα = √2cosθ 

Solution:

Given that tanθ = (sinα – cosα)/(sinα + cosα)

Now, on dividing both numerator and denominator by cosα, we get

⇒ tanθ = (tanα – 1)(tanα + 1)        

⇒ tanθ = (tanα – tan(π/4))(1+tan(π/4)tanα)

⇒ tanθ = tan(α – π/4)

⇒ θ = (α – π/4)  

Now Taking cos on both sides, we get  

⇒ cosθ = cos(α – π/4)   

⇒ cosθ = cosα.cos(π/4) + sinα.sin(π/4)

⇒ cosθ = cosα(1/√2) + sinα(1/√2)

⇒ cosθ = (cosα + sinα)/√2

⇒ √2cosθ = sinα + cosα

Hence Proved

Question 34. If tan(A + B) = p, tan(A – B) = q, then show that tan2A = (p + q)/(1 – pq)

Solution:

Given that, tan(A + B) = p, tan(A – B) = q

Now let’s solve RHS,

(p + q)/(1 – pq) = 

= 

= 

= 

= 

= 

= 

= tan2A = LHS

LHS = RHS

Hence Proved.