Tanalpha=3/4,secalpha=13/5 neither aplha beta is in 1st quadrant find cos(alpha-beta)?

Answer 1

Assuming you meant sec beta =13/5
#color(white)("XXX")cos(alpha-beta)=-56/65#

If #tan(alpha)=3/4# and #alpha# is not in Quadrant I
then #alpha# is an angle in Quadrant III (based on the CAST system, a positive value fro #tan# only occurs in Quadrants I and III.

Similarly, if #sec(beta)color(white)("xx")[=1/(cos(beta))]color(white)("xx")=13/5# and #beta# is not in Quadrant I
then #beta# is an angle in Quadrant IV.

We can draw these angles with appropriate right angled triangles and using the Pythagorean Theorem determine the length of the third side of each triangle (in the image below the calculated side's length is the circled value).

Using the #cos# angle-difference formula:
#color(white)("XXX")cos(alpha-beta)=cos(alpha)cos(beta) + sin(alpha)sin(beta)#

We can use the above images to evaluate this as
#color(white)("XXX")cos(alpha-beta)=((-4)/5) * (5/12) + ((-3)/5) *(12/13)#

#color(white)("XXX"cos(alpha-beta))=(-20)/65+ (-36)/65 = -56/65#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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