9) The angle A is found in quadrant 3, such that = −

. Determine the exact value
of sec (Decimal answers using a calculator will not be accepted). [3]
10) All progression work must be shown for full marks. Any unclear or questionable steps
made will not be credited.
Prove the identity: !"#$
#%!$ − !"&$
&'($ = − ) ∙ ) [4

Remember sin is positive on first and second quadrant cause it relates to y and cos is positive on first and fourth quadrant cause it relates to x.

From this use trig ratios and relations for finding others.

According to the given question

Angle A is found in 3rd quadrant such that sinA = -3/8

An image is attached of a right angled triangle

We know that

(Hypotenuse)² = (Opposite)² + (Base)²

Given

sinA = opposite/hypotenuse

∴ Opposite is -3 and hypotenuse is 8

8² = (-3)² + (base)²

base = [tex]\sqrt{64 - 9}[/tex]

base = [tex]\sqrt{55}[/tex]

We know that SecA = 1/cosA and cosA = Base/Hypotenuse

∴ SecA = Hypotenuse/Base

SecA = 8/[tex]\sqrt{55}[/tex] and sec is negative on the third quadrant.

So, in 3rd quadrant SecA= -8/[tex]\sqrt{55}[/tex].

learn more about Trigonometric functions here :

https://brainly.com/question/14746686

#SPJ1