Trigonometry question involving identities

[kalyan601]

Question:

It is required to solve the equation sinθcosθ = 1/4

i) Show that sinθ/cosθ + cosθ/sinθ = 1/sinθcosθ

Answer:
using the identity sin^2θ+cos^2θ=1,

sinθ/cosθ + cosθ/sinθ
= sin^2θ+cos^2θ/cosθsinθ
=1/sinθcosθ <- this is correct but I'm not sure on the next part

ii) Hence show that the equation sinθcosθ = 1/4 is equivalent to tanθ = 1/tanθ = 4

I know that I should use sinθ/cosθ = tanθ, but I'm not sure how

Thanks for the help

 

[HallsofIvy]

Question:

It is required to solve the equation sinθcosθ = 1/4

i) Show that sinθ/cosθ + cosθ/sinθ = 1/sinθcosθ

Answer:
using the identity sin^2θ+cos^2θ=1,

sinθ/cosθ + cosθ/sinθ
= sin^2θ+cos^2θ/cosθsinθ
=1/sinθcosθ <- this is correct but I'm not sure on the next part

ii) Hence show that the equation sinθcosθ = 1/4 is equivalent to tanθ = 1/tanθ = 4

I suspect you have copied this incorrectly- as given it is not true. What is true is that
\(\displaystyle tan(\theta)+ \frac{1}{tan(\theta)}= 4\)
If \(\displaystyle sin(\theta)cos(\theta)= frac{1}{4}\), \(\displaystyle \frac{1}{sin(\theta)cos(\theta)}= 4\)
so using part (i), \(\displaystyle \frac{sin(\theta)}{cos(\theta)}+ \frac{cos(\theta)}{sin(\theta)}= 4\).

I know that I should use sin

θ/cosθ = tanθ, but I'm not sure how

Thanks for the help

 

[kalyan601]

I suspect you have copied this incorrectly- as given it is not true. What is true is that
\(\displaystyle tan(\theta)+ \frac{1}{tan(\theta)}= 4\)
If \(\displaystyle sin(\theta)cos(\theta)= frac{1}{4}\), \(\displaystyle \frac{1}{sin(\theta)cos(\theta)}= 4\)
so using part (i), \(\displaystyle \frac{sin(\theta)}{cos(\theta)}+ \frac{cos(\theta)}{sin(\theta)}= 4\).

Yes sorry, I made a typo

Thanks for the help!