Question 816758
First of all, I assume that the second fraction's numerator is (1+sin(A)). If so, then please put multiple-term numerators (and denominators) in parentheses. If I am wrong then please re-post your question.<br>
{{{cos(A)/(1+sin(A))+(1+sin(A))/cos(A)=4}}}
Here's a solution that may be faster than most. For reasons that will become clear soon, multiply the first fraction by (1-sin(A)):
{{{(cos(A)/(1+sin(A)))((1-sin(A))/(1-sin(A)))+(1+sin(A))/cos(A)=4}}}
Simplifying...
{{{(cos(A)(1-sin(A)))/(1-sin^2(A))+(1+sin(A))/cos(A)=4}}}
{{{(cos(A)(1-sin(A)))/cos^2(A)+(1+sin(A))/cos(A)=4}}}
The factor of cos(A) in the numerator cancels one of the two factors of cos(A) in the denominator. (Do you see now why we multiplied by 1-sin(A)?)
{{{(1-sin(A))/cos(A)+(1+sin(A))/cos(A)=4}}}
The denominators are the same so we can add. The sin's cancel:
{{{2/cos(A)=4}}}
Multiplying both sides by cos(A):
{{{2=4cos(A)}}}
Dividing by 4:
{{{1/2=cos(A)}}}
We should recognize that 1/2 is a special angle value for cos. It tells us that the reference angle is 60 degrees. Since the 1/2 is positive and since cos is positive in the first and fourth quadrants we should get general solution equations of:
A = 60 + 360n (for the first quadrant)
A = -60 + 360n (for the fourth quadrant)<br>
Now we try various integers for n as we look for specific solutions within the given interval.
From A = 60 + 360n
if n = 0 then A = 60
if n = 1 (or larger) then A is too large for the interval
if n = -1 (or smaller) then A is too small for the interval
From A = -60 + 360n
if n = 0 (or smaller) then A is too small for the interval
if n = 1 then A = 300
if n = 2 (or larger) then A is too large for the interval<br>
So the only solutions within the given interval are: 60 and 300.