# 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

Taking LHS

= sin2Î¸cos2Î¸

Hence, LHS = RHS (Proved)

### Question 15. Prove that

Solution:

We Have

Taking LHS

= cosÎ¸/sinÎ¸

= cotÎ¸

Hence, LHS = RHS(Proved)

### Question 16. Prove that cosÎ¸(tanÎ¸ + 2)(2tanÎ¸ + 1) = 2secÎ¸ + 5sinÎ¸

Solution:

We have

cosÎ¸(tanÎ¸ + 2)(2tanÎ¸ + 1) = 2secÎ¸ + 5sinÎ¸

Taking LHS

= cosÎ¸(tanÎ¸ + 2)(2tanÎ¸ + 1)

= 2secÎ¸ + 5sinÎ¸

Hence, LHS = RHS(Proved)

Solution:

We have

x =

Taking LHS

=

### Question 18. If, then find the values of tanÎ¸, secÎ¸, and cosecÎ¸

Solution:

We have

As we know that

cosÎ¸ = âˆš1 – sin2Î¸         -(1)

Now put the value of sinÎ¸ in eq(1)

cosÎ¸ =

So the value of cosÎ¸ =

Now,

tanÎ¸ =

secÎ¸ =

cosecÎ¸ =

Alternative Method:

We have

We draw a â–³PQR 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) âˆ’ (a4 + b4 âˆ’ 2a2b2)

QR2 = 4a2b2

QR = 2ab

cosÎ¸ =

Now,

tanÎ¸ =

secÎ¸ =

cosecÎ¸ =

### Question 19.  If tanÎ¸ = a/b, then find the value of

Solution:

We have

=

Now put tanÎ¸ = a/b

### Question 20. If tanÎ¸ = a/b, show that .

Solution:

We have

Taking LHS

=

Dividing denominator and Numerator by cosÎ¸

=

=

=

=

=

=

Hence, LHS = RHS(Proved)

### Question 21. If cosecÎ¸ – sinÎ¸ = a3, secÎ¸ – cosÎ¸ = b3, then prove that a2b2(a2 + b2) = 1.

Solution:

Given: cosecÎ¸ – sinÎ¸ = a3

1/sinÎ¸ âˆ’ sinÎ¸ = a3

= a3

cos2Î¸/sinÎ¸ = a3

a = (cos2Î¸/sinÎ¸)1/3

Similarly, b = (sin2Î¸/cosÎ¸)1/3

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

Taking LHS

= a2b2(a2 + b2)

= a4b2 + a2b4

=

= cos6/3Î¸ + sin6/3Î¸

= cos2Î¸ + sin2Î¸

= 1

Hence, LHS = RHS (Proved)

### Question 22. If cotÎ¸(1 + sinÎ¸) = 4m and cotÎ¸(1 âˆ’ sinÎ¸) = 4n, prove that (m2 – n2)2 = mn.

Solution:

Given: cotÎ¸(1 + sinÎ¸) = 4m and cotÎ¸(1 âˆ’ sinÎ¸) = 4n

Multiplying both the equations

16mn = cot2Î¸(1 – sin2Î¸)

16mn =

16mn = cos4Î¸/sin2Î¸

mn = cos4Î¸/16sin2Î¸          -(1)

Now squaring the given equations

16m2 = cot2Î¸(1 + sinÎ¸)2 and 16n2 = cot2Î¸(1 – sinÎ¸)2

On subtracting both the equation, we get

16m2 – 16n2 = cot2Î¸(1 + sinÎ¸)2 – cot2Î¸(1 – sinÎ¸)2

16(m2 – n2) = cot2Î¸((1 + sinÎ¸)2 – (1 – sinÎ¸)2)

16(m2 – n2) =

(m2 – n2) = cos2Î¸/4sinÎ¸

On squaring both side, we get

(m2 – n2)2 = cos4Î¸/16sinÎ¸          -(2)

From equation(1) and (2)

(m2 – n2)2 = mn

Hence proved

### Question 23. If sinÎ¸ + cosÎ¸ = m then prove that sin6Î¸ + cos6Î¸ = , where m2 â‰¤ 2.

Solution:

Given: sinÎ¸ + cosÎ¸ = m

On squaring both side, we get

(sinÎ¸ + cosÎ¸)2 = m2

= sin2Î¸ + cos2Î¸ + 2sinÎ¸cosÎ¸ = m2

= 2sinÎ¸cosÎ¸ = m2 âˆ’ 1

Now,

Taking LHS

= sin6Î¸ + cos6Î¸

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

= (sin2Î¸)3 + (cos2Î¸)3

= (sin2Î¸ + cos2Î¸)(sin4Î¸ + cos4Î¸ âˆ’ sin2Î¸cos2Î¸)

= (1)((sin2Î¸)2 + (cos2Î¸)2 âˆ’ sin2Î¸cos2Î¸)

= (sin2Î¸ + cos2Î¸)2 âˆ’ 2sin2Î¸cos2Î¸ âˆ’ sin2Î¸cos2Î¸

= (1 âˆ’ 3sin2Î¸cos2Î¸)

Hence, Proved.

### Question 24. If a = secÎ¸ – tanÎ¸ and b = cosecÎ¸ + cotÎ¸, then show that ab + a – b + 1 = 0.

Solution:

We have

a = secÎ¸ – tanÎ¸ and b = cosecÎ¸ + cotÎ¸

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

= (secÎ¸ – tanÎ¸)(cosecÎ¸ + cotÎ¸) – (secÎ¸ – tanÎ¸) + (cosecÎ¸ + cotÎ¸) + 1

= (1/cosÎ¸ – sinÎ¸/cosÎ¸)(1/sinÎ¸ + cosÎ¸/sinÎ¸) – (1/cosÎ¸ – sinÎ¸/cosÎ¸) + (1/sinÎ¸ + cosÎ¸/sinÎ¸) + 1

= 1/cosÎ¸sinÎ¸ + 1/cosÎ¸ x cosÎ¸/sinÎ¸ – sinÎ¸/cosÎ¸ x 1/sinÎ¸ – (sinÎ¸/cosÎ¸) x (cosÎ¸/sinÎ¸) + 1/cosÎ¸ – sinÎ¸/cosÎ¸ – 1/sinÎ¸ – cosÎ¸/sinÎ¸ + 1

= 1/cosÎ¸sinÎ¸ + 1/sinÎ¸ – 1/cosÎ¸ – 1 + 1/cosÎ¸ – sinÎ¸/cosÎ¸ – 1/sinÎ¸ – cosÎ¸/sinÎ¸ + 1

= 1/cosÎ¸sinÎ¸ – sinÎ¸/cosÎ¸ – cosÎ¸/sinÎ¸

= 1 – sin2Î¸ – cos2Î¸/sinÎ¸cosÎ¸

= 1 – (sin2Î¸ + cos2Î¸)/sinÎ¸cosÎ¸

= 1 – 1/sinÎ¸cosÎ¸

= 0

Hence, LHS = RHS (Proved)

### Question 25. , where Ï€/2 < Î¸ < Ï€.

Solution:

We have

Taking LHS

= 2/cosÎ¸

Since Ï€/2 < Î¸ < Ï€   ,where cosÎ¸ is negative

So, -2/cosÎ¸

Hence, LHS = RHS (Proved)

Solution:

LHS =

= sin2Î¸cos2Î¸

RHS =

=

= sin2Î¸cos2Î¸

### 2T6 – 3T4 + 1 = 0

Solution:

LHS = 2(sin6Î¸ + cos6Î¸) – 3(sin4Î¸ + cos4Î¸) + 1

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

= 2(sin2Î¸ + cos2Î¸)(sin4Î¸ + cos4Î¸ – sin2Î¸cos2Î¸) – 3(sin4Î¸ + cos4Î¸) + 1

= 2(1)(sin4Î¸ + cos4Î¸ – sin2Î¸cos2Î¸) – 3(sin4Î¸ + cos4Î¸) + 1

= 2sin4Î¸ + 2cos4Î¸ – 2sin2Î¸cos2Î¸ – 3sin4Î¸ – 3cos4Î¸ + 1

= -sin4Î¸ – cos4Î¸ – 2sin2Î¸cos2Î¸ + 1

= -(sin2Î¸ + cos2Î¸)2 + 1

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

### 6T10 – 15T8 + 10T6 – 1 = 0

Solution:

T6 = sin6Î¸ + cos6Î¸

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

= (sin2x)3 + (cos2x)3

= (sin2x + cos2x)(sin4x + cos4x âˆ’ sin2xcos2x)

Using a2 + b2 = (a + b)2 âˆ’ 2ab

= (1)(sin4x + cos4x âˆ’ sin2xcos2x)

= (sin2x)2 + (cos2x)2 âˆ’ sin2xcos2x

= (sin2x + cos2x)2 âˆ’ 3sin2xcos2

= 1 âˆ’ 3sin2xcos2x

Similarly, we get the values of T8 & T10

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

= 1 âˆ’ 3sin2xcos2x âˆ’ sin2xcos2x(1 âˆ’ 2sin2xcos2x)

= 1 âˆ’ 4sin2xcos2x + 2sin4xcos4x

T10 = sin10Î¸ + cos10Î¸

= (sin6Î¸ + cos6Î¸)(sin4Î¸ + cos4Î¸) âˆ’ sin4Î¸cos4Î¸(sin2Î¸ + cos2Î¸)

= (1 âˆ’ 3sin2xcos2x)(1 âˆ’ 2sin2xcos2x) âˆ’ sin4xcos4x

= 1 âˆ’ 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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