Ex-5.1 Trigonometric Ratios (Part - 4), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

23.)   If 8tanA = 15 , find sinA–cosA

Sol. Given:

8tanA = 15

Therefore,

tanA = 15/8 …. (1)

To find:

sinA–cosA

Now we know tan A is defined as follows

tanA = Perpendicular side opposite to∠A/Base side adjacent to∠A …. (2)

Now by comparing equation (1) and (2)

We get

Perpendicular side opposite to ∠A  = 15

Base side adjacent to ∠A  = 8

Therefore triangle representing angle A is as shown below

Side AC = is unknown and can be found by using Pythagoras theorem

Therefore,

AC² = AB²  + BC²

Now by substituting the value of known sides from figure (a)

We get,

AC² = 15²  + 8²

AC²  = 225 + 64

AC = 289

Now by taking square root on both sides

We get,

AC =

AC = 17

Therefore Hypotenuse side AC = 17 …. (3)

Now we know, sin A is defined as follows

sinA = Perpendicular side opposite to∠AHypotenuse

Therefore from figure (a) and equation (3)

We get,

sinA = BC/AC

sinA = 15/17 …. (4)

Now we know, cos A is defined as follows

cosA = Base side adjacent to∠A/Hypotenuse

Therefore from figure (a) and equation (3)

We get,

cosA = AB/AC

cosA = 8/17 …. (5)

Now we find the value of expression sinA–cosA

Therefore by substituting the value the value of sinA and cosA from equation (4) and (5) respectively , we get,

sinA–cosA = 15/17–8/17

sinA–cosA = 

sinA–cosA = 7/17

Hence, sinA–cosA = 7/17

24.) If tanΘ = 20/21, show that 

Sol.
Given:

tanΘ = 20/21

To show that 

Now we know that

tanΘ = Perpendicular side opposite to∠Θ/Base side adjacent to∠Θ

Therefore,

tanΘ = 20/21

Side AC be the hypotenuse and can be found by applying Pythagoras theorem

Therefore,

AC²  = AB²  + BC²

AC² = 21²  + 20²

AC²  = 441 + 400

AC²  = 841

Now by taking square root on both sides

We get,

AC =  

AC = 29

Therefore Hypotenuse side AC = 29

Now we know, sinΘ is defined as follows,

sinAΘ = Perpendicular side opposite to∠Θ/Hypotenuse

Therefore from figure and above equation

We get,

sinΘ = AB/AC

sinΘ = 20/29

Now we know cosΘ is defined as follows

cosΘ = Base side adjacent to∠Θ/Hypotenuse

Therefore from figure and above equation

We get,

cosΘ = AB/AC

cosΘ = 21/29

Now we need to find the value of expression 

Therefore by substituting the value of sinΘ and cosΘfrom above equations, we get

Therefore after evaluating we get,

Hence,

25.) If cosecA = 2 , find 

Sol. Given:

cosecA = 2

To find 

Now cosec A =  Hypotenuse/Oppositeside  =  2/1

Here BC is the adjacent side,

By applying Pythagoras theorem,

AC² = AB²  + BC²

4 = 1 + BC²

BC² = 3

BC =  √3

Now we know that

sinA = 1/cosecA

sinA = 1/2 …. (1)

tanA = AB/BC

tanA =

cosA = BC/AC

cosA =

Substitute all the values of sinA , cosA and tanA from the equations(1) ,(2) and (3) respectively

We get.

= 2

Hence,

26.)   If ∠A and ∠B are acute angles such that cos A = cos B , then show that ∠A =  ∠B

Sol. Given:

∠A and ∠B are acute angles

cos A = cos B such that ∠A  =  ∠B

Let us consider right angled triangle ACB

Now since cos A = cos B

Therefore

AC/AB = BC/AB

Now observe that denominator of above equality is same that is AB

Hence  AC/AB = BC/AB only when AC = BC

Therefore AC = BC

We know that when two sides of triangle are equal, then opposite of the sides are also

Equal.

Therefore

We can say that

Angle opposite to side AC = angle opposite to side BC

Therefore,

∠B  =  ∠A

Hence, ∠A  =  ∠B

27.)  In a ΔABC , right angled triangle at A, if tan C =  √3 , find the value of sin B cos C + cos B sin C.

Sol. Given:

ΔABC

To find : sin B cos C + cos B sin C

The given a ΔABC is as shown in figure

Side BC is unknown and can be found using Pythagoras theorem,

Therefore,

BC² = AB²  + AC²

BC²  = 4

Now by taking square root on both sides

We get,

BC =  √4

BC = 2

Therefore Hypotenuse side BC = 2 …. (1)

Now, sin B =  Perpendicular side opposite to∠BHypotenuse

Therefore,

sinB = ACBC

Now by substituting the values from equation (1) and figure

We get,

sin B =  1/2 …. (2)

Now, cos B =  Base side adjacent to∠BHypotenuse

Therefore,

cos B =  AB/BC

Now substituting the value from equation

Similarly

Now by definition,

tanC = sinC/cosC

So by evaluating

cosC = 1/2 …. (5)

Now, by substituting the value of sinB, cosB,sin C and cosC from equation (2) ,(3) ,(4) and (5) respectively in sinB cosC + cosB sin C

sinB cosC + cosB sin C =  

=  1/4 + 3/4

= 1

Hence,

sinB cosC + cosB sin C = 1

28.)   State whether the following are true or false. Justify your answer.

(i)     The value od tan A is always less than 1.

(ii)    sec A =  12/5 for some value of ∠A.

(iii) cos A is the abbreviation used for the cosecant of ∠A.

(iv) sinΘ = 4/3 for some angle Θ.

Sol.

(i) tan A < 1

Value of tan A at 45^o i.e… tan  45 = 1

As value os A increases to 90^o

Tan A becomes infinite

So given statement is false.

(ii) sec A =  12/5 for some value of angle if

M-I

sec A = 2.4

sec A > 1

So given statements is true.

M- II

For sec A =  12/5 we get adjacent side = 13

Subtending 9i at B.

So, given statement is true.

(iii) Cos A is the abbreviation used for cosecant of angle A.

The given statement is false.

As such cos A is the abbreviation used for cos of angle A , not as cosecant of angle A.

(iv) Cot A is the product of cot A and A

Given statement is false

∵ cot A is a co-tangent of angle A and co-tangent of angle A =  adjacent side/Oposite side.

(v) sinΘ = 4/3 for some angle Θ.

Given statement is false

Since value of sinΘ is less than(or) equal to one.

Here value of sinΘ exceeds one,

So given statement is false.

29.) If sinΘ =   

Sol.

Given: sinΘ = 12/13

To Find:  

As shown in figure

Here BC is the adjacent side,

By applying Pythagoras theorem,

AC² = AB² + BC²

169 = 144 + BC²

BC² = 169 – 144

BC² = 25

BC = 5

Now we know that,

cosΘ = Base side adjacent to∠Θ/Hypotenuse

cosΘ = BC/AC

cosΘ = 5/13

We also know that,

tanΘ = sinΘ/cosΘ

Therefore, substituting the value of sinΘ and cosΘ from above equations

We get,

tanΘ = 12/5

Now substitute all the values of sinΘ , cosΘ and tanΘ from above equations in 

We get,

Therefore by further simplifying we get,

Therefore,

Hence,

30.)   If cosΘ = 5/13, find the value of 

Sol. Given: If cosΘ = 5/13

To find:

The value of expression 

Now we know that

cosΘ =  Base side adjacent to∠Θ/Hypotenuse …. (2)

Now when we compare equation (1) and (2)

We get,

Base side adjacent to ∠Θ  = 5

Hypotenuse = 13

Therefore, Triangle representing ∠Θ is as shown below

Perpendicular side AB is unknown and it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

AC² = AB²  + BC²

Therefore by substituting the values ogf known sides,

AB²  = 13² – 5²

AB² = 169 -25

AB²  = 144

AB = 12 …. (3)

Now we know from figure and equation,

sinΘ = 12/13 …. (4)

Now we know that,

tanΘ = sinΘ/cosΘ

tanΘ = 12/5 …. (5)

Now w substitute all the values from equation (1), (4) and (5) in the expression below,

Therefore

We get,

Therefore by further simplifying we get,

Therefore,

Hence,

31.)   If sec A =  17/8 , verify that 

Sol. Given:   sec A =  17/8

To verify:

Now we know that cosA = 1secA

Now, by substituting the value of sec A

We get,

cosA = 8/17

Now we also know that,

sin²A + cos²A = 1

Therefore

sin²A = 1–cos²A

=  (8/17)²

= 225/289

Now by taking square root on both sides,

We get,

sinA = 15/17

We also know that , tanA = sinA/cosA

Now by substituting the value of all the terms ,

We get,

tanA = 15/8

Now from the expression of above equation which we want to prove:

L.H.S =  

Now by substituting the value of cos A ad sin A from equation (3) and (4)

We get,

=  33/611

From expression

R.H.S =  frac³–tan²A1–3tan²A

Now by substituting the value of tan A from above equation

We get,

R.H.S =  

=

=  33/611

Therefore,

We can see that,

32.)  If sinΘ = 3/4, prove that 

Sol. Given: sinΘ = 3/4 …. (1)

To prove:

By definition,

sin A =  Perpendicular side opposite to∠A/Hypotenuse …. (3)

By comparing (1) and (3)

We get,

Perpendicular side = 3 and

Hypotenuse = 4

Side BC is unknown.

So we find BC by applying Pythagoras theorem to right angled ΔABC

Hence,

AC²  = AB²  + BC²

Now we substitute the value of perpendicular side (AB) and hypotenuse (AC) and get the base side (BC)

Therefore,

4² = 3²  + BC²

BC²  = 16 – 9

BC² = 7

BC =  √7

Hence, Base side BC = √7 …. (3)

Now cos A =  BC/AC

Now , cosecA = 1sinA

Therefore, from fig and equation (1)

cosecA = Hypotenuse/Perpendicular

cosecA = 4/3 …. (5)

Now, similarly

secA =

Further we also know that

cotA = cosA/sinA

Therefore by substituting th values from equation (1) and (4),

We get,

Now by substituting the value of cosec A, sec A and cot A from the equations  (5), (6), and (7) in the L.H.S of expression (2)

Hence it is proved that,

33.)  If secA = 17/8 , verify that 

Sol. Given: secA = 17/8 …. (1)

To verify:

Now we know that sec A =  1/cosA

Therefore cosA = 1/secA

We get,

cosA = 8/17 …. (3)

Similarly we can also get,

sin A =  sinA = 15/17 …. (4)

An also we know that tanA = sinAcosA

tanA = 15/8 …. (5)

Now from the expression of equation (2)

L.H.S: Missing close brace

Now by substituting the value of cos A and sin A from equation (3) and (4)

We get,

L.H.S =  

= 33/611 …. (6)

R.H.S =

Now by substituting the value of tan A from equation (5)

We get,

R.H.S =

= 33/611 …. (7)

Now by comparing equation (6) and (7)

We get,

34.) If cotΘ = 3/4, prove that 

Sol.

Given: cotΘ = 3/4

Prove that:

Now we know that

Here AC is the hypotenuse and we can find that by applying Pythagoras theorem

AC² = AB²  + BC²

AC²  = 16 + 9

AC² = 25

AC = 5

Similarly

secΘ = AC/BC

secΘ = 5/3

cosec = AC/AB

cosec = 5/4

Now on substituting the values in equations we get,

Therefore,

35.) If 3cosΘ–4sinΘ = 2cosΘ + sinΘ ,find tanΘ           

Sol.

Given: 3cosΘ–4sinΘ = 2cosΘ + sinΘ

To find: tanΘ

We can write this as:

3cosΘ–4sinΘ = 2cosΘ + sinΘ

cosΘ = 5sinΘ

Dividing both the sides by cosΘ ,

We get,

cosΘ/cosΘ = 5sinΘ/cosΘ

1 = 5tanΘ

tanΘ = 1

Hence,

 tanΘ = 1

36.) If ∠A and ∠P are acute angles such that tan A = tan P, then show ∠A = ∠P

Sol.

Given: A and P are acute angles tan A = tan P

Prove that: ∠A = ∠P

Let us consider right angled triangle ACP

We know tanΘ = oppositeside/adjacentside

tan A = PC/AC

tan P = AC/PC

∴ tan A = tan P

PC/AC = AC/PC

PC = AC [∵Angle opposite to equal sides are equal]

∠A = ∠P<