Question

If sin3A/sinA=k,then tan3A/tanA is:​

Answer 1

Answer:

see down

Step-by-step explanation:

Using tan3A=  

1−3tan  

2

A

3tanA−tan  

3

A

​  

 

We get  

tanA

tan3A

​  

=k=  

1−3tan  

2

A

3−tan  

2

A

​  

 

So, tan  

2

A=  

1−3k

k−3

​  

=  

1−sin  

2

A

sin  

2

A

​  

 

So, sin  

2

A=  

4(k−1)

k−3

​  

 

Now,  

sinA

sin3A

​  

=  

sinA

3sinA−4sin  

3

A

​  

=3−4sin  

3

A=3−  

1−k

k−3

​  

 

=  

k−1

2k

​  

 

Now, tan  

2

A>0  

So,  

3k−1

k−3

​  

>0  

So, k<  

3

1

​  

  or  k>3