# 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 Ï€/12 + tan Ï€/6 + tan Ï€/12 tan Ï€/6 = 1**

**(iii) tan 36**Â°** + tan 9Â° + tan 36Â° tan 9Â° = 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 2xProof:

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 Ï€/12 + tan Ï€/6 + tan Ï€/12 tan Ï€/6 = 1Proof:

As we know that

Ï€/12 15Â° and Ï€/6 = 30Â°

So, we have 15Â° + 30Â° = 45Â°

tan (15Â° +30Â°) = tan 45Â°

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

So,

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

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

After rearranging we get,

tan15Â° + tan30Â° + tan 15Â° tan30Â° = 1

Hence proved.

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

As we know that

36Â° + 9Â° = 45Â°

tan (36Â° + 9Â°) = tan 45Â°

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

So,

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

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

After rearranging we get,

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

LHS = RHS

Hence proved.

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

Let solve LHS,

= tan 13x – tan 9x -tan 4x

â‡’ 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,

= tan3Î¸ tanÎ¸

= tan(2Î¸ + Î¸) x tan(2Î¸ – Î¸)

=

=

= LHS

LHS = RHS

Hence proved

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

**Solution:**

Given that

â‡’

â‡’

Now by using componendo and Dividendo, we get

â‡’

â‡’

tanx/tany = a/bHence Proved.

### Question 20. If tanA = x tanB, prove that

**Solution:**

Given that

tanA = x tanB

sinA/cosA = x sinB/cosBâ‡’ 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) = m^{2} + n^{2} – 2

**Solution:**

Given that

cosA + sinB = m and sinA + cosB = n

Prove: 2sin(A + B) = m

^{2}+ n^{2}– 2Proof:

Let’s solve RHS, m

^{2}+ n^{2}– 2= (cosA + sinB)

^{2}+ (sinA + cosB)^{2}– 2= cos

^{2}A + sin^{2}B + 2cosA sinB + sin^{2}A + cos^{2}B + 2 sinA cosB – 2= (sin

^{2}A + cos^{2}A) + (sin^{2}B + cos^{2}B) + 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

â‡’ 1/tanA + 1/tanB = b

â‡’ (tanA + tanB)/(tanA tanB) = b

â‡’ a/(tanA tanB) = b

â‡’ 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 Î¸ lies in the first quadrant and cosÎ¸ = 8/17, then prove that:

### cos(Ï€/6 + Î¸) + cos(Ï€/4 – Î¸) + cos(2Ï€/3 – Î¸) = {(âˆš3 – 1)/2 + 1/âˆš2}23/17.

**Solution:**

Given,

0 < x < Ï€/2

Now, sinx =

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

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

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

= (8/17)(âˆš3/2 + 1/âˆš2 – 1/2) + (15/17)(-1/2 + 1/âˆš2 + âˆš3/2)

= (8/17)((âˆš3-1)/2 + 1/âˆš2) + (15/17)((âˆš3 – 1)/2 + 1/âˆš2)

= (23/17)((âˆš3-1)/2 + 1/âˆš2)

= RHS

LHS = RHS

Hence proved

### Question 25. tanx + tan(x + Ï€/3) + tan(x + 2Ï€/3) = 3, then prove that (3tanx – tan^{3}x)/(1 – 3tan^{2}x) = 1

**Solution:**

Given,

tanx + tan(x + Ï€/3) + tan(x + 2Ï€/3) = 3

Prove: (3tanx – tan

^{3}x)/(1 – 3tan^{2}x) = 1Proof:

â‡’

â‡’

â‡’

â‡’

â‡’

Hence proved.

### Question 26. If sin(Î± + Î²) = 1 and sin(Î± – Î²) = 1/2, where 0 â‰¤ Î±, Î² â‰¤ Ï€/2, then find the values of tan(Î± + 2Î²) and tan(2Î± + Î²)

** Solution:**

Given,

sin(Î± + Î²) = 1 and sin(Î± – Î²) = 1/2

Find the values of tan(Î± + 2Î²) and tan(2Î± + Î²)

So,

â‡’ Î± + Î² = 90Â° …..(i)

and Î± – Î² = 30Â° …..(ii)

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

â‡’ 2Î± = 120Â°

â‡’ Î± = 60Â°

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

â‡’ 2Î² = 60Â°

â‡’ Î² = 30Â°

So,

tan(Î± + 2Î²) = tan(60Â° + 2 Ã— 30Â°) = tan120Â° = -âˆš3

tan(2Î± + Î²) = tan(2 Ã— 60Â° + 30Â°) = tan150Â° = -1/âˆš3

### Question 27. If Î±, Î² are two different values of x lying between 0 and 2Ï€, which satisfy the equation 6cosx + 8sinx = 9, find the value of sin(Î± + Î²).

**Solution:**

Given,

6 cosx + 8 sinx = 9

â‡’ 6 cosx = 9 – 8 sinx

â‡’ 36 cos

^{2}x = (9 – 8 sinx)^{2}â‡’ 36(1 – sin

^{2}x) = 81 + 64sin^{2}x – 144 sinxâ‡’ 100 sin

^{2}x – 144 sinx + 45 = 0Now, let us considered Î± and Î² are the roots of the given equation,

So, cosÎ± and cosÎ² are the roots of the above equation.

â‡’ sinÎ± sinÎ² = 45/100

Again,

6cosx + 8sinx = 9

â‡’ 8sinx = 9 – 6 cosx

â‡’ 64 sin

^{2}x = (9 – 6 cosx)^{2}â‡’ 64(1 – cos

^{2}x) = 81 + 36 cos^{2}x – 108 cosxâ‡’ 100 cos

^{2}x – 108 cosx + 17 = 0Now, let us considered Î± and Î² are the roots of the given equation,

So, sinÎ± and sinÎ² are the roots of the above equation.

so, cosÎ± cosÎ² = 17/100

Hence, cos(Î± + Î²) = cosÎ± cosÎ² – sinÎ± sinÎ²

= 17/100 – 45/100

= -28/100

= -7/25

sin(Î± + Î²) = âˆš(1 – cos

^{2}(Î± + Î²))= âˆš(1 – (-7/25)

^{2})= âˆš(576/625

= 24/25

### Question 28 (i), If sinÎ± + sinÎ² = a and cosÎ± + cosÎ² = b, show that sin(Î± + Î²) = 2ab/(a^{2} + b^{2})

**Solution:**

Given that, sinÎ± + sinÎ² = a and cosÎ± + cosÎ² = b

Show : sin(Î± + Î²) = 2ab/(a

^{2}+ b^{2})So, now solve b

^{2}+ a^{2}= (cosÎ± + coÎ²)^{2}+ (sinÎ± + sinÎ²)^{2}= (cos

^{2}Î± + sin^{2}Î±) + (sin^{2}Î² + cos^{2}Î²) + 2(cosÎ± cosÎ² + sinÎ± sinÎ²)= 1 + 1 + 2 cos(Î± – Î²)

= 2 + 2 cos(Î± – Î²) ……..(i)

and,

b

^{2}– a^{2}= (cosÎ± + coÎ²)^{2}– (sinÎ± + sinÎ²)^{2}= cos

^{2}Î± + cos^{2}Î² – sin^{2}Î± – sin^{2}Î² + 2(cosÎ± cosÎ² – sinÎ± sinÎ²)= (cos

^{2}Î± – sin^{2}Î±) + (cos^{2}Î² – sin^{2}Î²) + 2 cos(Î± + Î²)= 2cos(Î± + Î²)cos(Î± – Î²) + 2cos(Î± + Î²)

= cos(Î± + Î²){2cos(Î± – Î²) + 2}

= cos(Î± + Î²)(b

^{2}+ a^{2}) …….(ii)â‡’ (b

^{2}– a^{2})/(b^{2}+ a^{2}) = cos(Î± + Î²)â‡’ sin(Î± + Î²) = âˆš(1 – cos

^{2}(Î± + Î²))=

= 2ab/(a

^{2}+ b^{2})

### Question 28 (ii). If sinÎ± + sinÎ² = a and cosÎ± + cosÎ²= b, show that cos(Î± + Î²) = (b^{2} – a^{2})/(b^{2} + a^{2})

**Solution:**

Given that, sinÎ± + sinÎ² = a and cosÎ± + cosÎ²= b

Show: cos(Î± + Î²) = (b

^{2}– a^{2})/(b^{2}+ a^{2})So, now solve b

^{2}+ a^{2}= (cosÎ± + coÎ²)^{2}+ (sinÎ± + sinÎ²)^{2}= (cos

^{2}Î± + sin^{2}Î±) + (sin^{2}Î² + cos^{2}Î²) + 2(cosÎ± cosÎ² + sinÎ± sinÎ²)= 1 + 1 + 2 cos(Î± – Î²)

= 2 + 2 cos(Î± – Î²) ……(i)

and,

b

^{2}– a^{2}= (cosÎ± + coÎ²)^{2}– (sinÎ± + sinÎ²)^{2}= cos

^{2}Î± + cos^{2}Î² – sin^{2}Î± – sin^{2}Î² + 2(cosÎ± cosÎ² – sinÎ± sinÎ²)= (cos

^{2}Î± – sin^{2}Î±) + (cos^{2}Î² – sin^{2}Î²) – 2 cos(Î± + Î²)= 2cos(Î± + Î²) cos(Î± – Î²) + 2cos(Î± – Î²)

= cos(Î± + Î²) {2cos(Î± – Î²) + 2} ……..(ii)

Now from (i) and (ii), we have

â‡’ b

^{2}– a^{2}= cos(Î± + Î²)(a^{2}+ b^{2})â‡’ (b

^{2}– a^{2})/(b^{2}+ a^{2}) = cos(Î± + Î²)

### 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 sinÎ± sinÎ² – cosÎ± cosÎ² + 1 = 0, proved that 1 + cotÎ± tanÎ² = 0

**Solution:**

Given,

sinÎ± sinÎ² – cosÎ± cosÎ² + 1 = 0

â‡’ -(cosÎ± cosÎ² – sinÎ± sinÎ²) + 1 = 0

â‡’ -cos(Î± + Î²) + 1 = 0

â‡’ cos(Î± + Î²) = 1

Therefore, sin(Î± + Î²) = 0 ……(i)

Let’s solve LHS

= 1 + cotÎ± tanÎ² = 1 + (cosÎ± sinÎ²)/(sinÎ± cosÎ²)

= (sinÎ± cosÎ² + cosÎ± sinÎ²)/ (sinÎ± cosÎ²)

= sin(Î± + Î²)/ (sinÎ± cosÎ²)

Now from eq(i), we get

= 0

LHS = RHS

Hence Proved.

### Question 31. tanÎ± = x + 1 and tanÎ² = x – 1, show that 2cot(Î± – Î²) = x^{2}

**Solution:**

We have,

tanÎ± = x + 1 and tanÎ² = x – 1

As we know that tan(Î± – Î²) = (tanÎ± – tanÎ²) / (1 + tanÎ± tanÎ²)

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

= (x + 1 – x + 1) / (1 + x

^{2}– 1)= 2/ (1 + x

^{2 }– 1)= 2/x

^{2}cot(Î± – Î²) = x

^{2}/22cot(Î± – Î²) = x

^{2}LHS = RHS

Hence Proved.

### Question 32. If angle Î¸ is divided into two parts such that the tangents of one part is Î» part times the tangent of the other, and Ï• is their difference then show that sinÎ¸ = (Î» + 1)/(Î» – 1) sinÏ•.

**Solution:**

Let us considered Î± and Î² be the two parts of the angle be Î¸.

Then, Î¸ = Î± + Î² and Ï• = Î± – Î²

According to question, we get

tanÎ± = Î» tanÎ²

â‡’ tanÎ± / tanÎ² = Î»/1

Now, applying componendo and dividendo, we get

â‡’ (tanÎ± + tanÎ²) / (tanÎ± – tanÎ²) = (Î»+1) / (Î»-1)

â‡’

â‡’

â‡’

â‡’

â‡’

Hence proved.

### Question 33. If tanÎ¸ = (sinÎ± – cosÎ±)/(sinÎ± + cosÎ±), then show that sinÎ± + cosÎ± = âˆš2cosÎ¸

**Solution:**

Given that tanÎ¸ = (sinÎ± – cosÎ±)/(sinÎ± + cosÎ±)

Now, on dividing both numerator and denominator by cosÎ±, we get

â‡’ tanÎ¸ = (tanÎ± – 1)(tanÎ± + 1)

â‡’ tanÎ¸ = (tanÎ± – tan(Ï€/4))(1+tan(Ï€/4)tanÎ±)

â‡’ tanÎ¸ = tan(Î± – Ï€/4)

â‡’ Î¸ = (Î± – Ï€/4)

Now Taking cos on both sides, we get

â‡’ cosÎ¸ = cos(Î± – Ï€/4)

â‡’ cosÎ¸ = cosÎ±.cos(Ï€/4) + sinÎ±.sin(Ï€/4)

â‡’ cosÎ¸ = cosÎ±(1/âˆš2) + sinÎ±(1/âˆš2)

â‡’ cosÎ¸ = (cosÎ± + sinÎ±)/âˆš2

â‡’ âˆš2cosÎ¸ = sinÎ± + cosÎ±

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.

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