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

Chapter 7 of RD Sharma's Class 11 Mathematics textbook focuses on "Trigonometric Ratios of Compound Angles." This chapter delves into the trigonometric identities related to the compound angles, offering a deeper understanding of how angles interact within the trigonometric functions. It provides essential formulas and identities crucial for solving complex trigonometric problems.

Trigonometric Ratios of Compound Angles

The trigonometric ratios of compound angles involve the expressions that include the sum or difference of two angles. These ratios help simplify trigonometric expressions and solve equations involving compound angles. Key formulas include:

• Sum of Angles: sin(A+B)=sinAcosB+cosAsinB
• Difference of Angles: cos(A−B)=cosAcosB+sinAsinB
• Double Angle Formulas: sin2A=2sinAcosA and cos2𝐴=cos²𝐴−sin²𝐴

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 \frac{tan^22θ-tan^2θ}{1-tan^22θtan^2θ}=tan3θtanθ

Solution:

Prove:  \frac{tan^22θ-tan^2θ}{1-tan^22θtan^2θ}=tan3θtanθ

Proof:

Le's solve RHS,

= tan3θ tanθ

= tan(2θ + θ) x tan(2θ - θ)

= [\frac{tan2θ+tanθ}{1-tan2θtanθ}][\frac{tan2θ-tanθ}{1+tan2θtanθ}]

= \frac{tan^22θ-tan^2θ}{1-tan^22θtan^2θ}

= LHS

LHS = RHS

Hence proved

Question 19. If \frac{sin(x+y)}{sin(x-y)}=\frac{a+b}{a-b}   , show that tanx/tany = a/b

Solution:

Given that \frac{sin(x+y)}{sin(x-y)}=\frac{a+b}{a-b}

⇒ \frac{sinxcosy+sinycosx}{sinxcosy-sinycosx}=\frac{a+b}{a-b}

⇒ \frac{sinxcosy+sinycosx+sinxcosy-sinycosx}{sinxcosy+sinycosx-sinxcosy+sinycosx}=\frac{a+b+a-b}{a+b-a+b}

Now by using componendo and Dividendo, we get

⇒ \frac{2sinxcosy}{2sinycosx}=\frac{2a}{2b}

⇒ tanx/tany = a/b

Hence Proved.

Question 20. If tanA = x tanB, prove that \frac{sin(A-B)}{sin(A+B)}=(x-1)/(x+1)

Solution:

Given that

tanA = x tanB

 sinA/cosA = x sinB/cosB

⇒ sinAcosB = x cosA sinB

Now, \frac{sin(A-B)}{sin(A+B)}=\frac{sinAcosB-sinBcosA}{sinAcosB+cosAsinB}

= \frac{xcosAsinB-cosAsinB}{xcosAsinB+cosAsinB}

= \frac{cosAsinB(x-1)}{cosAsinB(x+1)}

= (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)]

= \frac{tan(A+B)+tan(A-B)}{1-tan(A+B)tan(A-B)}

= (x + y) / (1 - xy)

Since, tan2B = tan[(A + B) - (A - B)]

So, 

= \frac{tan(A+B)-tan(A-B)}{1+tan(A+B)tan(A-B)}

= (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 = \sqrt{1-cos^2x}=\sqrt{1-64/289}=15/17

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:

⇒ tanx + \frac{tanx+tan\frac{π}{3}}{1-tanx.tan\frac{π}{3}}+\frac{tanx+tan\frac{2π}{3}}{1-tanx.tan\frac{2π}{3}}=3

⇒ tanx+\frac{tanx+√3}{1-√3tanx}+\frac{tanx-√3}{1+√3tanx}=3

⇒ \frac{tanx(1-3tan^2x)+tanx+√3+√3tan^2x+3tanx+tanx-√3-√3tan^2x+3tanx}{1-3tan^2x}=3

⇒ \frac{9tanx-3tan^3x}{1-3tan^2x}=3

⇒ \frac{3tanx-tan^3x}{1-3tan^2x}=1

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))

= \sqrt{1-(\frac{b^2-a^2}{b^2+a^2})^2}=\sqrt{\frac{b^4+a^4-b^4-a^4+4a^2b^2}{(b^2+a^2)^2}}

= 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 \frac{1}{sin(x-a)sin(x-b)}=\frac{cot(x-a)-cot(x-b)}{sin(a-b)}

Solution:

Let's solve RHS

= \frac{cot(x-a)-cot(x-b)}{sin(a-b)}

= \frac{\frac{cos(x-a)}{sin(x-a)}-\frac{cos(x-b)}{sin(x-b)}}{sin(a-b)}

= \frac{sin(x-b)cos(x-a)-sin(x-a)cos(x-b)}{sin(x-a)sin(x-b)sin(a-b)}

= \frac{sin(x-b-x+a)}{sin(x-a)sin(x-b)sin(a-b)}

= \frac{sin(a-b)}{sin(x-a)sin(x-b)sin(a-b)}

= \frac{1}{sin(x-a)sin(x-b)}

= LHS

LHS = RHS

Hence proved.

Question 29 (ii). Proved that \frac{1}{sin(x-a)cos(x-b)}=\frac{cot(x-a)+tan(x-b)}{cos(a-b)}

Solution:

Let's solve RHS

= \frac{cot(x-a)+tan(x-b)}{cos(a-b)}

= \frac{\frac{cos(x-a)}{sin(x-a)}+\frac{sin(x-b)}{cos(x-b)}}{cos(a-b)}

= \frac{cos(x-b)cos(x-a)+sin(x-a)sin(x-b)}{cos(x-b)sin(x-a)cos(a-b)}

= \frac{cos(x-b-x+a)}{sin(x-a)cos(x-b)cos(a-b)}

= \frac{cos(a-b)}{sin(x-a)cos(x-b)cos(a-b)}

= \frac{1}{sin(x-a)cos(x-b)}

= RHS

LHS = RHS

Hence Proved.

Question 29 (iii). Proved that \frac{1}{cos(x-a)cos(a-b)}=\frac{tan(x-b)-tan(x-a)}{sin(a-b)}

Solution:

Let's solve RHS

= \frac{tan(x-b)-tan(x-a)}{sin(a-b)}

= \frac{\frac{sin(x-b)}{cos(x-b)}-\frac{sin(x-a)}{cos(x-a)}}{sin(a-b)}

= \frac{sin(x-b)cos(x-a)-sin(x-a)sin(x-b)}{cos(x-b)cos(x-a)sin(a-b)}

= \frac{sin(x-b-x+a)}{cos(x-a)cos(x-b)sin(a-b)}

= \frac{sin(a-b)}{cos(x-a)cos(x-b)sin(a-b)}

= \frac{1}{cos(x-a)cos(x-b)}

= 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)          

⇒ \frac{\frac{sinα}{cosα}+\frac{sinβ}{cosβ}}{\frac{sinα}{cosα}-\frac{sinβ}{cosβ}}=\frac{λ+1}{λ-1}

⇒ \frac{\frac{sinαcosβ+cosαsinβ}{cosαcosβ}}{\frac{sinαcosβ-cosαsinβ}{cosαcosβ}}=\frac{λ+1}{λ-1}

⇒ \frac{sin(α+β)}{sin(α-β)}=\frac{λ+1}{λ-1}

⇒ \frac{sinθ}{sinϕ}=\frac{λ+1}{λ-1}

⇒ sinθ=\frac{λ+1}{λ-1}sinϕ

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) = \frac{tan(A+B)+tan(A-B)}{1-tan(A+B).tan(A-B)}

= \frac{\frac{tanA+tanB}{1-tanA.tanB}+\frac{tanA-tanB}{1+tanA.tanB}}{1-\frac{tanA+tanB}{1-tanA.tanB}.\frac{tanA-tanB}{1+tanA.tanB}}

= \frac{\frac{(tanA+tanB)(1+tanAtanB)+(tanA-tanB)(1-tanAtanB)}{(1-tanAtanB)(1+tanAtanB)}}{\frac{(1-tanAtanB)(1+tanAtanB)-(tanA+tanB)(tanA-tanB)}{(1-tanAtanB)(1+tanAtanB)}}

= \frac{tanA+tanB+tan^2AtanB+tanAtan^2B+tanA-tanB-tan^2AtanB+tanAtan^2B}{1-tan^2Atan^2B-tan^2A+tan^2B}

= \frac{2tanA+2tanA.tan^2B}{(1-tan^2A)(1+tan^2B)}

= \frac{2tanA(1+tan^2B)}{(1-tan^2A)(1+tan^2B)}

= \frac{2tanA}{1-tan^2A}

= tan2A = LHS

LHS = RHS

Hence Proved.

Read More:

• Trigonometric Ratios
• Trigonometric Ratios of Some Specific Angles