Question

3 cosece - 4 cote = 0 then find the value of sine, cos e.​

Answer 1

Answer:

Let △PQR be a right angled triangle where ∠Q=90  

0

 and ∠R=A as shown in the above figure:

Now it is given that 4sinA−3cosA=0 that is  

4sinA=3cosA

⇒  

cosA

sinA

​

=  

4

3

​

 

⇒tanA=  

4

3

​

 

We know that, in a right angled triangle, tanθ is equal to opposite side over adjacent side that is tanθ=  

Adjacentside

Oppositeside

​

, therefore, opposite side PQ=3 and adjacent side QR=4.

Now, using pythagoras theorem in △PQR, we have

PR  

2

=PQ  

2

+QR  

2

=3  

2

+4  

2

=9+16=25

⇒PR=  

25

​

=5

Therefore, the hypotenuse PR=5.

We know that, in a right angled triangle,

sinθ is equal to opposite side over hypotenuse that is sinθ=  

Hypotenuse

Oppositeside

​

 and

 

cosθ is equal to adjacent side over hypotenuse that is cosθ=  

Hypotenuse

Adjacentside

​

 

Here, we have opposite side PQ=3, adjacent side QR=4 and the hypotenuse PR=5, therefore, the trignometric ratios of angle A can be determined as follows:

sinA=  

Hypotenuse

Oppositeside

​

=  

PR

PQ

​

=  

5

3

​

 

cosA=  

Hypotenuse

Adjacentside

​

=  

PR

QR

​

=  

5

4

​

 

cosec A=  

sinA

1

​

=  

5

3

​

 

1

​

=1×  

3

5

​

=  

3

5

​

 

secA=  

cosA

1

​

=  

5

4

​

 

1

​

=1×  

4

5

​

=  

4

5

​

 

 

Hence, sinA=  

5

3

​

, cosA=  

5

4

​

, cscA=  

3

5

​

 and secA=  

4

5

​

.

Step-by-step explanation: