To find the value of tan x² given that sin x = 35 and x is in the third quadrant, we first need to correct the value of sin x. Since the sine function can only have values between -1 and 1, it’s likely there was a mistake in the value provided. Assuming a more realistic scenario, let’s assume this was meant to be a sine of an angle in the third quadrant.
First, we know that in the third quadrant, both sine and cosine are negative. Thus, we can express the sine in the following way:
We need to find cos x using the identity:
sin²x + cos²x = 1
Since sin x can be represented as a negative value in the third quadrant, we take sin x = - rac{3}{5} (as a valid example), then:
sin²x + cos²x = 1
(- rac{3}{5})² + cos²x = 1
rac{9}{25} + cos²x = 1
cos²x = 1 – rac{9}{25} = rac{16}{25}
Thus, cos x = - rac{4}{5} (since cosine is also negative in the third quadrant).
Now we can find tan x which is defined as:
tan x = sin x / cos x
With our values:
tan x = rac{-3/5}{-4/5} = rac{3}{4}
Next, we want to find tan x²:
tan²x = (tan x)² = ( rac{3}{4})² = rac{9}{16}
As a result, the value of tan x² is rac{9}{16}.