Question

If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] 

Options

• \[\frac{16}{625}\]

• \[\frac{1}{36}\]

• \[\frac{3}{160}\]

• \[\frac{160}{3}\]

Answer 1

Given: cos θ = 3/5 and we need to find the value of the following expression` "sinθ tanθ-1"/"2tan^2 θ"` 

We know that `cos θ = "Base"/"Hypotenuse"` 

⇒`"Base"=3 `

⇒ `"Hypotenuse"=5` 

⇒`" Perpendicular"= sqrt(("Hypotenuse")^2-("Base")^2)` 

⇒ `"Perpendicular"= sqrt(25-9)` 

⇒`"Perpendicular"=4`

`"Since" sin θ= "Perpendicular"/"Hypotenuse"` 

and tan θ= `"Perpendicular"/"Base" ` 

So we find, 

`(sin θ tan θ-1)/(2 tan^2 θ)` 

`(4/5xx4/3-1)/(2xx(4/3)^2)` 

`(16/15-1)/(32/9)` 

`(1/15)/(32/9)`

`3/160` 

Hence the correct option is (c78)