Given sin x =-4/5 and x is in quadrent 3, what is the value of tan x/2

We can write sin x in terms of tan x/2 using the formula:

⇒ sin x = (2 tan (x/2)) / (1 + tan2(x/2))

Therefore, using the above formula, we can find the values of tan x/2 by putting the value of sin x.

⇒ -4/5 = (2 tan (x/2)) / (1 + tan2(x/2))

Now, if we replace tan (x/2) by y, we get a quadratic equation:

⇒ 0.8y2 + 2y + 0.8 = 0

⇒ 2y2 + 5y + 2 = 0

By using the quadratic formula, we get y = -0.5, -2

Hence, the value of tan (x/2) = -0.5, -2

Now, we have two solutions of tan (x/2).

Now, let's check for the ideal solution using the formula tan x = (2 tan (x/2)) / (1 - tan2(x/2)).

For tan (x/2) = -0.5:

⇒ tan x = 2(-0.5) / 1 - (-0.5)2 = -4/3

It is also given that x lies in the third quadrant. We know that tan is positive in the third quadrant, and here we get tan x = -4/3 which is negative.

Hence, we can say that tan (x/2) = -0.5 is not a correct solution. Hence it is rejected.

Now let's check for tan (x/2) = -2.

⇒ tan x = 2(-2) / 1 - (-2)2 = 4/3

Here, we get tan x = 4/3 which is positive.

Hence, we can say that tan (x/2) = -2 is a correct solution.

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