Ex-5.1 Trigonometric Ratios (Part - 3), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

12.) If tanΘ = a/b, prove that 

Sol.

Given:

tanΘ  a/b …. (1)

Now, we know that tanΘ = sinΘ/cosΘ

Therefore equation (1) becomes

sinΘ/cosΘ = a/b ….(2)

Now, by multiplying by a/b on both sides of equation (2)

We get,

a/b × sinΘ/cosΘ = a/b × a/b

Therefore,

asinΘ/bcosΘ = a²/b² ….(3)

Now by applying dividendo in above equation (3)

We get,

Now by applying componendo in equation (3)

We get,

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

We get,

Therefore,

Therefore, bcosΘ and b² cancels on L.H.S and R.H.S respectively

Hence, it is proved that

13.)   If secΘ = 13/5, show that 

 Sol.

Given:

secΘ = 13/5

To show that 

Now, we know that cosΘ = 1/secΘ

Therefore,

cosΘ =

Therefore,

cosΘ = 5/13 …. (1)

Now, we know that

cosΘ =

Now, by comparing equation (1) and(2)

We get,

Base side adjacent to ∠Θ  = 5

And

Hypotenuse = 13

Therefore from above figure

Base side BC = 5

Hypotenuse AC = 13

Side AB is unknown. It can be determined by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

AC²  = AB²  + BC²

Therefore by substituting the values of known sides

We get,

13²  = AB²  + 5²

Therefore,

AB² = 13² – 5²

AB² = 169 – 25

AB²  = 144

AB =  

Therefore,

AB = 12 …. (3)

Now, we know that

sinΘ = AB/AC

sinΘ = 12/13 …. (4)

Now L.H.S of the equation to be proved is as follows

L.H.S =  

Substituting the value cosΘ of sinΘand from equation (1) and (4) respectively

We get,

Therefore,

L.H.S =  9/3

L.H.S = 3

Hence proved that,

14.)   If  cosΘ = 12/13 , show that sinΘ(1–tanΘ) = 35/156

Sol. Given: cosΘ = 12/13 …. (1)

To show that sinΘ(1–tanΘ) = 35/156

Now we know that  cosΘ = Basesideadjacentto∠Θ/Hypotenuse ….(2)

Therefore, by comparing equation (1) and (2)

We get,

Base side adjacent to ∠Θ  = 12

And

Hypotenuse = 13

Therefore from above figure

Base side BC = 12

Hypotenuse AC = 13

Side AB is unknown and it can be determined by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

AC² = AB²  + BC²

Therefore by substituting the values of known sides

We get,

13² = AB²  + 12²

Therefore,

AB² = 13²– 12²

AB² = 169 – 144

AB = 25

AB =  

Therefore,

AB = 5 …. (3)

Now, we know that

sinΘ = Perpendicularsideoppositeto∠Θ/Hypotenuse

Now from figure (a)

We get,

sinΘ = AB/AC

Therefore,

sinΘ = 5/12 …. (5)

Now L.H.S of the equation to be proved is as follows

L.H.S of the equation to be proved is as follows

L.H.S =  sinΘ(1–tanΘ] …. (6)

Substituting the value of sinΘ and tanΘ from equation (4) and (5)

We get,

Taking L.C.M inside the bracket

We get,

Therefore,

Now by opening the bracket and simplifying

We get,

L.H.S =  35/136

From equation (6) and (7) ,it can be shown that

that sinΘ(1–tanΘ)  =  35/136

15.) If cotΘ = , show that 

Sol. Given: cotΘ = …. (1)

To show that 

Now, we know that cotΘ = 1/tanΘ

Since tanΘ = Perpendicular side oppositeto∠Θ/Base side adjacentto∠Θ

Therefore,

tanΘ =

Therefore,

cotΘ = Base side adjacentto∠Θ/Perpendicular side oppositeto∠Θ …. (2)

Comparing Equation (1) and (2)

We get.

Base side adjacent to ∠Θ  = 1

Perpendicular side opposite to ∠Θ  =  √3

Therefore, triangle representing angle √3 is as shown below

Therefore, by substituting the values of known sides

We get,

AC²  =  (√3)²  + 1²

Therefore,

AC²  = 3 + 1

AC² = 4

AC =  √4

Therefore,

AC = 2 …. (3)

Now, we know that

sinΘ = Perpendicular side oppositeto∠Θ/Hypotenuse

Now from figure (a)

sinΘ = AB/AC

Therefore from figure (a) and equation (3),

sinΘ =

Now we know that

Now from figure (a)

We get,

BC/AC

Therefore from figure (a) and equation (3),

cosΘ = 1/2 …. (5)

Now, L.H.S of the equation to be proved is as follows

L.H.S =  

Substituting the value of from equation (4) and (5)

We get,

L.H.S =

L.H.S =  

Now by taking L.C.M in numerator as well as denominator

We get,

L.H.S =

Therefore,

L.H.S =  

L.H.S =  3/4 × 4/5

L.H.S =  3/5  = R.H.S

Therefore,

16.)   If tanΘ = , then show that 

Sol. Given: tanΘ =  …. (1)

To show that     

Now, we know that

Since, tanΘ = Perpendicular side oposite to∠Θ/Base side adjacent to∠Θ ….(2)

Therefore,

Comparing equation (1) and (2)

We get.

Perpendicular side opposite to ∠Θ  = 1

Base side adjacent to ∠Θ  =  √7

Therefore, Triangle representing ∠Θ is shown below

Hypotenuse AC 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 of known sides

We get,

AC² = (1)²  +  (√7)²

Therefore,

AC ²  = 1 + 7

AC²  = 8

AC =  √8

AC =  

Therefore,

AC =  2√2 …. (3)

Now we know that

sinΘ = Perpendicular side opposite to∠Θ/Hypotenuse

sinΘ = AB/AC

sinΘ =

Now, we know that cosec Θ = 1/sinΘ

Therefore, from equation (4)

We get,

cosec Θ = 2√2  …. (5)

Now, we know that

cosΘ = Base side adjacent to∠Θ/Hypotenuse

Now from figure (a)

We get,

cosΘ = BC/AC

Therefore from figure (a) and equation (3)

cosΘ = 

Now we know that secΘ = 1/cosΘ

Therefore, from equation (6)

We get,

secΘ =

secΘ = 

Now, L.H.S of the equation to be proved is as follows

L.H.S =  

Substituting the value of cosecΘ andsecΘ from equation (6) and (7)

We get,

L.H.S =

L.H.S =

Therefore,

L.H.S =  

Therefore,

L.H.S =  48/64

L.H.S =  3/4  = R.H.S

Therefore,

Hence proved that

17.)    If secΘ = 5/4,find the value of 

Sol. Given: secΘ = 5/4 …. (1)

To find the value of 

Now we know that secΘ = 1/cosΘ

Therefore,

cosΘ = 1/secΘ

Therefore from equation (1)

cosΘ = 1/5

cosΘ = 4/5 …. (2)

Also, we know that cos²Θ + sin²Θ = 1

Therefore,

sin²Θ = 1–cos²Θ

sinΘ = 

Substituting the value of cosΘ from equation (2)

We get,

sinΘ =

=  9/25

=  3/5

Therefore,

sinΘ = 3/5 …. (3)

Also, we know that

sec²Θ = 1 + tan²Θ

Therefore,

tan²Θ = (5/4)² −1

tanΘ = 

Therefore,

tanΘ = 3/4 …. (4)

Also, cotΘ = 1/tanΘ

Therefore from equation (4)

We get,

cotΘ = 4/3 …. (5)

Substituting the value of cosΘ, cotΘ and tanΘ from the equation (2),(3),(4) and (5) respectively in the expression below

We get,

= 12/7

Therefore, 

18.)   If sinΘ = 12/13 , find the value of 

Sol. Given: sinΘ = 12/13 …. (1)

To, find the value of 

Now, we know the following trigonometric identity

cosec² Θ = 1 + tan²Θ

Therefore, by substituting the value of tanΘ from equation (1)

We get,

cosec² Θ = 1 + (12/13)²

By taking L.C.M on the R.H.S

We get,

cosec²Θ =

=  313/169

Therefore

cosecΘ =

Therefore

cosecΘ  =  Θ =

Now, we know that

cosecΘ  =  1/sinΘ

Therefore

Now, we know the following trigonometric identity

cos²Θ + sin²Θ = 1

Therefore,

cos²Θ = 1–sin²Θ

Now by substituting the value of sinΘ from equation (3)

We get,

= 1–169/313

Therefore, by taking L.C.M on R.H.S

We get,

cos²Θ = 144/313

Now, by taking square root on both sides

We get,

Therefore,

Substituting the value of sinΘ and cosΘ from equation (3) and (4) respectively in the equation below

Therefore,

312/25

Therefore

  312/25
 

19.) If cosΘ = 3/5 , find the value of 

Sol.

Given: cosΘ = 3/5 …. (1)

To find the value of 

Now we know the following trigonometric identity

cos²Θ + sin²Θ = 1

Therefore by substituting the value of cosΘ from equation (1)

We get,

(3/5)² + sin²Θ = 1

Therefore,

Therefore by taking square root on both sides

We get,

sinΘ = 4/5 …. (2)

Now, we know that

tanΘ = sinΘ/cosΘ

Therefore by substituting the value of sinΘ and cosΘ from equation (2) and (1) respectively

We get,

Now, by substituting the value of sinΘ and of tanΘ from equation (2) and equation (4) respectively in the expression below

We get,

Therefore,

20.)   If sinΘ = 35 ,  find the value of 

Sol. Given:

sinΘ = 3/5 …. (1)

To find the value of 

Now, we know the following trigonometric identity

cos²Θ + sin²Θ = 1

Therefore by substituting the value of cosΘ from equation (1)

We get,

cos²Θ + (3/5)² = 1

Therefore,

cos²Θ = 1–(3/5)²

cos²Θ = 1–9/25

Now by taking L.C.M

We get,

Therefore, by taking square roots on both sides

We get,

cosΘ = 4/5

Therefore,

cosΘ = 4/5 …. (2)

Now we know that

tanΘ = sinΘ/cosΘ

Therefore by substituting the value of sinΘ and cosΘ from equation (1) and (2) respectively

We get,

tanΘ = 3/4 …. (3)

Also, we know that

cotΘ = 1/tanΘ

Therefore from equation (3)

We get,

cotΘ = 4/3 …. (4)

Now by substituting the value of cosΘ, tanΘ and cotΘ from equation (2) ,(3) and (4) respectively from the expression  below

We get,

Therefore, 
 

21.)   If tanΘ = 24/7, find that sinΘ + cosΘ

Sol.

Given:

tanΘ = 24/7 …. (1)

To find,

sinΘ + cosΘ

Now we know that tanΘ is defined as follows

tanΘ = Perpendicular side opposite to∠Θ/Base side adjacentto∠Θ …. (2)

Now by comparing equation (1) and (2)

We get,

Perpendicular side opposite to ∠Θ  = 24

Base side adjacent to ∠Θ  = 7

Therefore triangle representing ∠Θ 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 unknown sides from figure

We get,

AC² = 24²  + 7²

AC = 576 + 49

AC = 625

Now by taking square root on both sides,

We get,

AC = 25

Therefore H hy

Hypotenuse side AC = 25 …. (3)

Now we know sinΘ is defined as follows

sinΘ = Perpendicular side opposite to∠Θ/Hypotenuse

Therefore from figure (a) and equation (3)

We get,

sinΘ = AB/AC

sinΘ = 24/25 …. (4)

Now we know that cosΘ is defined as follows

cosΘ = Base side adjacent to∠Θ/Hypotenuse

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

sinΘ + cosΘ  =  24/25 + 7/25

sinΘ + cosΘ  =  31/25

Hence,  sinΘ + cosΘ  =  31/25

22.) If sinΘ = a/b, find secΘ + tanΘ in terms of a and b.

Sol. Given:

sinΘ = a/b …. (1)

To find: secΘ + tanΘ

Now we know, sinΘ is defined as follows

sinΘ = Perpendicularsideoppositeto∠ΘHypotenuse    …. (2)

Now by comparing equation (1) and (2)

We get,

Perpendicular side opposite to ∠Θ  = a

Hypotenuse = b

Therefore triangle representing ∠Θ is as shown below

Hence side BC is unknown

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

Therefore,

AC²  = AB²  + BC²

Now by substituting the value of sides AB and AC from figure (a)

We get,

b²  = a²  + BC²

Therefore,

BC²  = b² – a²

Now by taking square root on both sides

We get,

Therefore,

Base side  …. (3)

Now we know cosΘ is defined as follows

cosΘ = Base side adjacent to∠Θ/Hypotenuse

Therefore from figure (a) and equation (3)

We get,

cosΘ = BC/AC

=

cosΘ = BC/AC

Now we know, secΘ = 1/cosΘ

Therefore,

Now we know, tanΘ = sinΘ/cosΘ

Now by substituting the values from equation (1) and (3)

We get,

Therefore,

Now we need to find secΘ + tanΘ

Now by substituting the values of secΘ and tanΘ from equation (5) and (6) respectively

We get,

secΘ + tanΘ  =

secΘ + tanΘ  =

We get,

secΘ + tanΘ  =  

Now by substituting the value in above expression

We get,

secΘ + tanΘ   =

Now,  present in the numerator as well as denominator of above denominator of above expression gets cancels we get,

secΘ + tanΘ = 

Square root is present in the numerator as well as denominator of above expression

Therefore we can place both numerator and denominator under a common square root sign

Therefore, secΘ + tanΘ =