# Class 11 RD Sharma Solutions – Chapter 5 Trigonometric Functions – Exercise 5.1 | Set 2

• Last Updated : 21 Feb, 2021

### Question 14. Prove that Solution:

We have

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Hence, LHS = RHS (Proved)

### Question 15. Prove that Solution:

We Have Taking LHS     = cosθ/sinθ

= cotθ

Hence, LHS = RHS(Proved)

### Question 16. Prove that cosθ(tanθ + 2)(2tanθ + 1) = 2secθ + 5sinθ

Solution:

We have

cosθ(tanθ + 2)(2tanθ + 1) = 2secθ + 5sinθ

Taking LHS

= cosθ(tanθ + 2)(2tanθ + 1)      = 2secθ + 5sinθ

Hence, LHS = RHS(Proved)

### Question 17. If x = , prove that is also equal to x.

Solution:

We have

x = Taking LHS   =  ### Question 18. If , then find the values of tanθ, secθ, and cosecθ

Solution:

We have As we know that

cosθ = √1 – sin2θ         -(1)

Now put the value of sinθ in eq(1)

cosθ =     So the value of cosθ = Now,

tanθ = secθ = cosecθ = Alternative Method:

We have We draw a △PQR right-angled at Q PR = a2 + b2 and PQ = a2 – b2 By Pythagoras theorem, we have

PR2 = PQ2 + QR2

QR2 = (a2 + b2)2 – (a2 – b2)2

QR2 = (a4 + b4 + 2a2b2) − (a4 + b4 − 2a2b2)

QR2 = 4a2b2

QR = 2ab

cosθ = Now,

tanθ = secθ = cosecθ = ### Question 19.  If tanθ = a/b, then find the value of Solution:

We have = Now put tanθ = a/b     ### Question 20. If tanθ = a/b, show that .

Solution:

We have Taking LHS

= Dividing denominator and Numerator by cosθ

= = = = = = Hence, LHS = RHS(Proved)

### Question 21. If cosecθ – sinθ = a3, secθ – cosθ = b3, then prove that a2b2(a2 + b2) = 1.

Solution:

Given: cosecθ – sinθ = a3

1/sinθ − sinθ = a3 = a3

cos2θ/sinθ = a3

a = (cos2θ/sinθ)1/3

Similarly, b = (sin2θ/cosθ)1/3

Now putting the values of a and b in the following equation

Taking LHS

= a2b2(a2 + b2)

= a4b2 + a2b4

= = cos6/3θ + sin6/3θ

= cos2θ + sin2θ

= 1

Hence, LHS = RHS (Proved)

### Question 22. If cotθ(1 + sinθ) = 4m and cotθ(1 − sinθ) = 4n, prove that (m2 – n2)2 = mn.

Solution:

Given: cotθ(1 + sinθ) = 4m and cotθ(1 − sinθ) = 4n

Multiplying both the equations

16mn = cot2θ(1 – sin2θ)

16mn = 16mn = cos4θ/sin2θ

mn = cos4θ/16sin2θ          -(1)

Now squaring the given equations

16m2 = cot2θ(1 + sinθ)2 and 16n2 = cot2θ(1 – sinθ)2

On subtracting both the equation, we get

16m2 – 16n2 = cot2θ(1 + sinθ)2 – cot2θ(1 – sinθ)2

16(m2 – n2) = cot2θ((1 + sinθ)2 – (1 – sinθ)2)

16(m2 – n2) = (m2 – n2) = cos2θ/4sinθ

On squaring both side, we get

(m2 – n2)2 = cos4θ/16sinθ          -(2)

From equation(1) and (2)

(m2 – n2)2 = mn

Hence proved

### Question 23. If sinθ + cosθ = m then prove that sin6θ + cos6θ = , where m2 ≤ 2.

Solution:

Given: sinθ + cosθ = m

On squaring both side, we get

(sinθ + cosθ)2 = m2

= sin2θ + cos2θ + 2sinθcosθ = m2

= 2sinθcosθ = m2 − 1

Now,

Taking LHS

= sin6θ + cos6θ

Using a3 + b3 = (a + b)(a2 + b2 − ab)

= (sin2θ)3 + (cos2θ)3

= (sin2θ + cos2θ)(sin4θ + cos4θ − sin2θcos2θ)

= (1)((sin2θ)2 + (cos2θ)2 − sin2θcos2θ)

= (sin2θ + cos2θ)2 − 2sin2θcos2θ − sin2θcos2θ

= (1 − 3sin2θcos2θ)   Hence, Proved.

### Question 24. If a = secθ – tanθ and b = cosecθ + cotθ, then show that ab + a – b + 1 = 0.

Solution:

We have

a = secθ – tanθ and b = cosecθ + cotθ

and we have to proof that

ab + a – b + 1 = 0

So, taking LHS

ab + a – b + 1

Now put the values of a and b, we get

= (secθ – tanθ)(cosecθ + cotθ) – (secθ – tanθ) + (cosecθ + cotθ) + 1

= (1/cosθ – sinθ/cosθ)(1/sinθ + cosθ/sinθ) – (1/cosθ – sinθ/cosθ) + (1/sinθ + cosθ/sinθ) + 1

= 1/cosθsinθ + 1/cosθ x cosθ/sinθ – sinθ/cosθ x 1/sinθ – (sinθ/cosθ) x (cosθ/sinθ) + 1/cosθ – sinθ/cosθ – 1/sinθ – cosθ/sinθ + 1

= 1/cosθsinθ + 1/sinθ – 1/cosθ – 1 + 1/cosθ – sinθ/cosθ – 1/sinθ – cosθ/sinθ + 1

= 1/cosθsinθ – sinθ/cosθ – cosθ/sinθ

= 1 – sin2θ – cos2θ/sinθcosθ

= 1 – (sin2θ + cos2θ)/sinθcosθ

= 1 – 1/sinθcosθ

= 0

Hence, LHS = RHS (Proved)

### Question 25. , where π/2 < θ < π.

Solution:

We have Taking LHS       = 2/cosθ

Since π/2 < θ < π   ,where cosθ is negative

So, -2/cosθ

Hence, LHS = RHS (Proved)

### \frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_5}

Solution:

LHS =     = sin2θcos2θ

RHS = =     = sin2θcos2θ

### 2T6 – 3T4 + 1 = 0

Solution:

LHS = 2(sin6θ + cos6θ) – 3(sin4θ + cos4θ) + 1

Using (a3 + b3) = (a + b)(a2 + b2 – ab)

= 2(sin2θ + cos2θ)(sin4θ + cos4θ – sin2θcos2θ) – 3(sin4θ + cos4θ) + 1

= 2(1)(sin4θ + cos4θ – sin2θcos2θ) – 3(sin4θ + cos4θ) + 1

= 2sin4θ + 2cos4θ – 2sin2θcos2θ – 3sin4θ – 3cos4θ + 1

= -sin4θ – cos4θ – 2sin2θcos2θ + 1

= -(sin2θ + cos2θ)2 + 1

= -1 + 1 = 0 = RHS (Hence Proved)

### 6T10 – 15T8 + 10T6 – 1 = 0

Solution:

T6 = sin6θ + cos6θ

Using a3 + b3 = (a + b)(a2 + b2 − ab)

= (sin2x)3 + (cos2x)3

= (sin2x + cos2x)(sin4x + cos4x − sin2xcos2x)

Using a2 + b2 = (a + b)2 − 2ab

= (1)(sin4x + cos4x − sin2xcos2x)

= (sin2x)2 + (cos2x)2 − sin2xcos2x

= (sin2x + cos2x)2 − 3sin2xcos2

= 1 − 3sin2xcos2x

Similarly, we get the values of T8 & T10

T8 = (sin6x + cos6x)(sin2x + cos2x) − sin2xcos2x(sin4x + cos4x)

= 1 − 3sin2xcos2x − sin2xcos2x(1 − 2sin2xcos2x)

= 1 − 4sin2xcos2x + 2sin4xcos4x

T10 = sin10θ + cos10θ

= (sin6θ + cos6θ)(sin4θ + cos4θ) − sin4θcos4θ(sin2θ + cos2θ)

= (1 − 3sin2xcos2x)(1 − 2sin2xcos2x) − sin4xcos4x

= 1 − 5sin2xcos2x + 5sin4xcos4x

On putting the values of T6, T8 and T10 in the following equation

6T10 – 15T8 + 10T6 – 1

We get the value 0.

Hence Proved

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