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

Question 14. Prove that 

Solution:

We have

Taking LHS

= 

= 

= 

= 

= 

= 

= sin²θcos²θ

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 andPQ = 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

=

= cos^6/3θ + sin^6/3θ

= 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)

Question 26 (i). If T_n = sinⁿθ + cosⁿθ, prove that

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

Solution: 

LHS = 

= sin²θcos²θ

RHS = 

=

= sin2θcos²θ

Question 26 (ii). If T_n = sinⁿθ + cosⁿθ, prove that

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)

Question 26 (iii). If T_n = sinⁿθ + cosⁿθ, prove that

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