SOLUTION: Find the exact value of tan[cos^-1(1/4)-csc^-1(-3/2)]

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Question 627375: Find the exact value of tan[cos^-1(1/4)-csc^-1(-3/2)]
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
First of all, we should recognize that the expression "exact value" means "put away your calculator and solve this by hand". (If "exact value" was not used then we could just use our calculators and get a quick decimal approximation.)

An exponent of -1 on the name of a function indicates an inverse function. Inverse trig functions take a ratio as input and return the angle within a specific range which would have that ratio. For example:
cos%5E%28-1%29%281%2F4%29 would return the one and only angle between zero and pi which has a cos ratio of 1/4. And csc%5E%28-1%29%28-3%2F2%29 would return the one and only angle between -pi%2F2 and pi%2F2 whose csc ratio is -3/2. (Note how the two inverse functions have different ranges. Some have ranges between 0 and pi and the other have a range between -pi%2F2 and pi%2F2. When you work with these inverse Trig functions you have to learn which range goes with which inverse function.)

In summary, the inverse Trig functions return angles as output. So your expression:
tan%28cos%5E%28-1%29%281%2F4%29-csc%5E%28-1%29%28-3%2F2%29%29
represents the tan of the difference/subtraction of two angles. We have a formula for this:
tan%28A-B%29+=+%28tan%28A%29-tan%28B%29%29%2F%281%2Btan%28A%29%2Atan%28B%29%29
This formula tells how we can find that tan of a difference of angles using the tan's of the individual angles. So to find the exact value of your expression we will need the tan's of each the two given angles.

Let's start with tan%28cos%5E%28-1%29%281%2F4%29%29. The range of inverse cos is 0 to pi and within this range only angles between 0 and pi%2F2 are positive. So we know that the angle in question is between 0 and pi%2F2 (i.e. it is a 1st quadrant angle). The tan of a first quadrant angle will be positive. But which positive number? For this we can use a reference (right) triangle. Pick one of the acute angles to be the angle we are working with. We know from cos%5E%28-1%29%281%2F4%29 that the cos ratio is 1/4. Since cos is adjacent/hypotenuse we will use 1 for the adjacent side and 4 for the hypotenuse. To find the tan ratio we need opposite/adjacent. We have the adjacent side. We need the opposite side. Use the Pythagorean Theorem to find the opposite side. You should get sqrt%2815%29. So
tan%28cos%5E%28-1%29%281%2F4%29%29+=+sqrt%2815%29%2F1+=+sqrt%2815%29

We can use a similar process for tan%28csc%5E%28-1%29%28-3%2F2%29%29. The range for inverse csc is -pi%2F2 to pi%2F2. Our angle has a negative csc, -3/2. Within the range, only angles between pi%2F2 and 0 (i.e. a 4th quadrant angle) will have a negative csc. Fourth quadrant angles will have negative tan's. But which negative number? Again we can use a reference triangle. (Note: We already know we are supposed to get a negative result. So we can ignore signs in our triangle. csc is hypotenuse/opposite. So our hypotenuse is 3 and our opposite side is 2. tan is still opposite/adjacent so we will use the Pythagorean Theorem again to find the missing side. We should get sqrt%285%29 for the adjacent side. So
tan%28csc%5E%28-1%29%28-3%2F2%29%29+=+-2%2Fsqrt%285%29
(Note: if this was the final answer we should rationalize the denominator. But this is not our final answer and I prefer to postpone issues like rationalizing until the very end. (Sometimes these issues work themselves out on their own.))

Now that we have values for tan%28cos%5E%28-1%29%281%2F4%29%29 and tan%28csc%5E%28-1%29%28-3%2F2%29%29 we are ready to use our tan(A-B) formula on:
tan%28cos%5E%28-1%29%281%2F4%29-+csc%5E%28-1%29%28-3%2F2%29%29
From the formula we get:

Substituting in our values we get:

Simplifying we get:
%28sqrt%2815%29+%2B+2%2Fsqrt%285%29%29%2F%281-%282%2Asqrt%2815%29%29%2Fsqrt%285%29%29
%28sqrt%2815%29+%2B+2%2Fsqrt%285%29%29%2F%281-2%2Asqrt%283%29%29
We can eliminate the fraction within a fraction by multiplying the numerator and denominator by sqrt%285%29:

%28sqrt%2875%29+%2B+2%29%2F%28sqrt%285%29-2%2Asqrt%2815%29%29
Now we can deal with rationalizing. To rationalize a two-term denominator we multiply the top and bottom by the denominators conjugate: %28sqrt%285%29%2B2%2Asqrt%2815%29%29:

On top we must use FOIL to multiply. On the bottom we can use the %28a%2Bb%29%28a-b%29+=+a%5E2-b%5E2 pattern to multiply more quickly than FOIL:

Simplifying:

%28sqrt%28375%29%2B2sqrt%281175%29+%2B+2sqrt%285%29+%2B+4sqrt%2815%29%29%2F%285+-+60%29
%28sqrt%28375%29%2B2sqrt%281175%29+%2B+2sqrt%285%29+%2B+4sqrt%2815%29%29%2F%28-55%29
Last of all we simplify, if possible, the square roots that remain in the numerator. The first two have perfect square factors:


%285sqrt%2815%29%2B2%2A5%2A3%2Asqrt%285%29+%2B+2sqrt%285%29+%2B+4sqrt%2815%29%29%2F%28-55%29
%285sqrt%2815%29%2B30sqrt%285%29+%2B+2sqrt%285%29+%2B+4sqrt%2815%29%29%2F%28-55%29
And finally, adding like terms:
%289sqrt%2815%29%2B32sqrt%285%29%29%2F%28-55%29
or
-%289sqrt%2815%29%2B32sqrt%285%29%29%2F55