Question

If sin θ = `3/5`, tan φ = `1/2 and pi/2` < θ < π < φ < `(3pi)/2,`, then find the value of 8 tan θ – `sqrt5` sec φ.

Answer 1

Given that sin θ = `3/5 = "Opposite side"/"Hypotenuse"`

∵ AB = `sqrt(5^2 - 3^2)`

`= sqrt(25 - 9) = sqrt 16` = 4

Here θ lies in second quadrant [∵ `pi/2` < θ < π]

∵ tan θ is negative.

tan θ = `- 3/4`

Also given that tan Φ = `1/2 = "Opposite side"/"Adjacent side"`

∴ PR = `sqrt("PQ"^2 + "QP"^2)`

`= sqrt(4 + 1) = sqrt5`

Here Φ lies in third quadrant `(∵ π < Φ < (3pi)/2)`

∴ sec Φ is negative.

`sec phi = 1/(cos phi) = - 1/(2/sqrt3) = - sqrt5/2`

Now 8 tan θ – `sqrt5` sec Φ = `8(- 3/4) - sqrt5 ((- sqrt5)/2)`

= 2 × (-3) + `5/2`

`= - 6 + 5/2`

`= (-12+5)/2`

`= (-7)`