Ex-5.1 Trigonometric Ratios(Part - 2), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

 3.) In the below figure, find tan P and cot R. Is tan P = cot R? 

 To find, tan P, cot R

Sol.

In the given right angled ΔPQR, length of side OR is unknown

Therefore , by applying Pythagoras theorem in ΔPQR

We get,

PR²  = PQ² + QR²

Substituting the length of given side PR and PQ in the above equation

13² = 12² + QR²

QR²  = 13² – 12²

QR²  = 169 – 144

QR² = 25

QR =  

By definiton, we know that ,

tanP = Perpendicularsideoppositeto∠P/Basesideadjacentto∠P

tanP = QR/PQ

tanP = 5/12 …. (1)

Also, by definition, we know that

cotR = Basesideadjacentto∠R/Perpendicularsideoppositeto∠R

cotR = QR/PQ

cotR = 5/12 …. (2)

Comparing equation (1) ad (2), we come to know that that R.H.S of both the equation are equal.

Therefore, L.H.S of both equations is also equal

tan P = cot R

Answer:

Yes , tan P = cot R =  5/12

4.)  If sin A =  9/41, Compute cos A and tan A.

Sol. Given:  sinA = 9/41 …. (1)

To find: cos A, tan A

By definition,

sinA = Perpendicularsideoppositeto∠A/Hypotenuse …. (2)

By comparing (1) and (2)

We get ,

Perpendicular side = 9 and

Hypotenuse = 41

Now using the perpendicular side and hypotenuse we can construct ΔABC as shown below

Length of side AB is unknown is right angled ΔABC ,

To find the length of side AB, we use Pythagoras theorem,

Therefore, by applying Pythagoras theorem in ΔABC ,

We get,

AC²  = AB² + BC²

41²  = AB² + 9²

AB²  = 41² – 9²

AB²  = 168 – 81

AB = 1600

AB =  

AB = 40

Hence, length of side AB = 40

Now

By definition,

cosA = Basesideadjacentto∠A/Hypotenuse

cosA = AB/AC

cosA = 40/41

Now,

By definition,

tanA = Perpendicularsideoppositeto∠A/Basesideadjacentto∠A

tanA = BC/AB

tanA = 9/40

Answer:

cosA = 40/41 , tanA = 9/40

5.)  Given 15cot A = 8, find sin A and sec A.

Answer: Given: 15cot A = 8

To find: sin A, sec A

Since 15 cot A = 8

By taking 15 on R.H.S

We get,

cotA = 8/15

By definition,

cotA = 1/tanA

Hence,

cotA = 1Perpendicularsideoppositeto∠A/Basesideadjacentto∠A

cotA = Basesideadjacentto∠A/Perpendicularsideoppositeto∠A …. (2)

Comparing equation (1) and (2)

We get,

Base side adjacent to ∠A  = 8

Perpendicular side opposite to ∠A  = 15

ΔABC can be drawn below using above information

Hypotenuse side is unknown.

Therefore, we find side AC of ΔABC by Pythagoras theorem.

So, by applying Pythagoras theorem to ΔABC

We get,

AC²  = AB² +BC²

Substituting values of sides from the above figure

AC²  = 8² + 15²

AC²  = 64 + 225

AC²  = 289

AC =

AC = 17

Therefore, hypotenuse = 17

Now by definition,

sinA = Perpendicularsideoppositeto∠A/Hypotenuse

Therefore, sinA = BC/AC

Substituting values of sides from the above figure

sinA = 1517

By definition,

secA = 1/cosA

Hence,

secA =

secA = Hypotenuse/Basesideadjacentto∠A

Substituting values of sides from the above figure

secA = 17/8

Answer:

 sinA = 15/17, secA = 17/8

  6.) In ΔPQR , right angled at Q, PQ = 4cm and RQ = 3 cm .Find the value of sin P,   sin R , sec P and sec R.

Sol. Given:

ΔPQR is right angled at vertex Q.

PQ = 4cm

RQ = 3cm

To find,

sin P, sin R , sec P , sec R  

                Given ΔPQR is as shown below

Hypotenuse side PR is unknown.

Therefore, we find side PR of ΔPQR by Pythagoras theorem

By applying Pythagoras theorem to ΔPQR

We get,

PR²  = PQ² +RQ²

Substituting values of sides from the above figure

PR²  = 4² +3²

 PR²  = 16 + 9

PR²  = 25

PR =  

PR = 5

Hence, Hypotenuse = 5

Now by definition,

sinP = Perpendicularsideoppositeto∠P/Hypotenuse

sinP = RQ/PR

Substituting values of sides from the above figure

sinP = 3/5

Now by definition,

sinR = Perpendicularsideoppositeto∠R/Hypotenuse

sinR = PQ/PR

Substituting the values of sides from above figure

sinR = 4/5

By definition,

secP = 1/cosP

secP =

secP = Hypotenuse/Basesideadjacentto∠P

Substituting values of sides from the above figure

secP = PR/PQ

secP = 5/4

By definition,

secR = 1/cosR

secR =

secR = Hypotenuse/Basesideadjacentto∠R

Substituting values of sides from the above figure

secR = PR/RQ

secR = 5/3

Answer:

sinP = 3/5 , sinR = 4/5,

secP = 5/4, secR = 5/3

7.)   If cotΘ = 7/8, evaluate

(ii)  cot2Θ

Sol. Given: cotΘ = 7/8

To evaluate:  

We know the following formula

(a + b)(a – b) = a² – b²

By applying the above formula in the numerator of equation (1)

We get,

(1+sinθ)×(1–sinθ) = 1–sin2θ.....(2)(Where,a = 1andb = sinθ)

Similarly,

By applying formula (a +b) (a – b) = a² – b² in the denominator  of equation (1).

We get,

(1+cosΘ)(1–cosΘ) = 1²–cos²Θ … (Where a = 1 and b =   cosΘ

(1+cosΘ)(1–cosΘ) = 1–cos²Θ … (Where a = 1 and b =   cosΘ

Substituting the value of numerator and denominator of equation (1) from equation (2), equation (3).

Therefore,

Since,

cos²Θ + sin²Θ = 1

Therefore,

cos²Θ = 1–sin² Θ

Also, sin²Θ = 1–cos²Θ

Putting the value of 1–sin²Θ and 1–cos²Θ in equation (4)

We get,

We know that, 

Since, it is given that cotΘ = 7/8

Therefore,

(ii) Given: cotΘ = 7/8

To evaluate: cot²Θ

cotΘ = 7/8

Squaring on both sides,

We get,

(cotΘ)² = (7/8)²

(cotΘ)² = 49/64

Answer:

49/64

8.) If 3cotA = 4 , check whether   = cos²A – sin²A or not.

Sol. Given: 3cot A = 4

To check whether  = cos²A – sin²A or not.

3cot A = 4

Dividing by 3 on both sides,

We get,

cot A =  4/3 …. (1)

By definition,

cotA = 1/tanA

Therefore,

cotA =

cotA = Basesideadjacentto∠A/Perpendicularsideoppositeto∠A …. (2)

Comparing (1) and (2)

We get,

Base side adjacent to ∠A  = 4

Perpendicular side opposite to ∠A  = 3

Hence ΔABC is as shown in figure below

In ΔABC , Hypotenuse is unknown

Hence, it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem inΔABC

We get

AC² = AB² +BC²

Substituting the values of sides from the above figure

AC²  = 4² + 3²

AC²  = 16 +9

AC²  = 25

AC =

AC = 5

Hence, hypotenuse = 5

To check whether

To check whether  = cos²A–sin²A or not.

We get thee values of tan A , cos A , sin A

By definition,

tan A =  1/cotA

Substituting the value of cot A from equation (1)

We get,

tan A  =  1/4

tan A =  3/4 …. (3)

Now by definition,

cosA = Basesideadjacentto∠A/Hypotenuse

cosA = AB/AC

Substituting the values of sides from the above figure

cosA = 4/5 …. (5)

Now we first take L.H.S of equation  = cos²A – sin²A

L.H.S =

Substituting value of tan A from equation (3)

We get,

Taking L.C.M on both numerator and denominator

We get,

Now we take R.H.S of equation whether  = cos²A–sin²A

R.H.S =  cos²A – sin²A

Substituting value of sin A and cos A from equation (4) and (5)

We get,

R.H.S =

Comparing (6) and (7)

We get.

 = cos²A–sin²A

Answer:

Yes, = cos²A–sin²A

 9.)  If tanΘ = ab , find the value of 

Sol.

Given:

tanΘ = a/b …. (1)

Now, we know that tanΘ = sinΘ/cosΘ

Therefore equation (1) become as follows

sinΘ/cosΘ  = a/b

Now, by applying invertendo

We get,

cosΘ/sinΘ = b/a

Now by applying Componendo – dividendo

We get,

Therefore,

10.)    If 3tanΘ = 4 , find the value of 

Sol.

Given: If 3tanΘ = 4

Therefore,

tanΘ = 4/3 …. (1)

Now, we know that tanΘ = sinΘ/cosΘ

Therefore equation (1) becomes

sinΘ/cosΘ = 4/3 ….(2)

Now, by applying Invertendo to equation (2)

We get,

cosΘ/sinΘ = 3/4 …. (3)

Now, multiplying by 4 on both sides

We get

4×cosΘ/sinΘ = 4×3/4

Therefore

Now, multiplying by 2 on both sides of equation (3)

We get,

2cosΘ/sinΘ = 3/2

Now by applying componendo in above equation

We get,

Therefore,

Therefore, on L.H.S sinΘ cancels and we get,

Therefore,

4cosΘ–sinΘ = 4

11.)   If 3cotΘ = 2, find the value of                                                                                                                                                                                     Sol.

Given:

3cotΘ = 2

Therefore,

cotΘ = 2/3 …. (1)

Now, we know that cotΘ = cosΘ/sinΘ

Therefore equation (1) becomes

cosΘ/sinΘ = 2/3 ….(2)

Now , by applying invertendo to equation (2)

sinΘ/cosΘ = 3/2 ….(3)

Now, multiplying by 43 on both sides,

We get,

4/3 × sinΘ/cosΘ = 4/3 × 3/2

Therefore, 3 cancels out on R.H.S and

We get,

4sinΘ/3cosΘ = 2/1

Now by applying invertendo dividendo in above equation

We get,

Now, multiplying by 2/6 on both sides of equation (3)

We get,

2/6 × sinΘ/cosΘ = 2/6 × 3/2

Therefore, 2 cancels out on R.H.S and

We get,

2sinΘ/6cosΘ = 3/6

2sinΘ/6cosΘ = 1/2

Now by applying componendo in above equation

We get,

Now, by dividing equation (4) by (5)

We get,

Therefore,

Therefore, on L.H.S (3 sinΘ) cancels out and we get,

Now, by taking 2 in the numerator of L.H.S on the R.H.S

We get,

Therefore, 2 cancels out on R.H.S and

We get,

Hence answer,