Minimizing 4sec2(x)+9csc2(x) for x in the first quadrant. Discrepancy in solutionUsing derivatives, I am able to show that the minimum value of the expression-in-question is equal to 25. I also verified this with Desmos graphing app. However, when I tried doing this using basic algebra, the answer turns out to be 26. I do not know why this is happening, but I should be extremely grateful to you if you can point out the error.4sec2x+9csc2x=(2sec⁡x−3csc⁡x)2+12sec⁡xcsc⁡xThe value of the above expression will be minimum when that expression inside parenthesis equals zero; and that happens when tan⁡x=32. Using this we can say that sec⁡x=132 and csc⁡x=133If now we substitute these values in the above expression, answer turns out to be 26. I don't know why this is happening, but please help me find the error in this approach.

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Minimizing 4sec2(x)+9csc2(x) for x in the first quadrant. Discrepancy in solutionUsing derivatives, I am able to show that the minimum value of the expression-in-question is equal to 25. I also verified this with Desmos graphing app. However, when I tried doing this using basic algebra, the answer turns out to be 26. I do not know why this is happening, but I should be extremely grateful to you if you can point out the error.4sec2x+9csc2x=(2sec⁡x−3csc⁡x)2+12sec⁡xcsc⁡xThe value of the above expression will be minimum when that expression inside parenthesis equals zero; and that happens when tan⁡x=32. Using this we can say that sec⁡x=132 and csc⁡x=133If now we substitute these values in the above expression, answer turns out to be 26. I don&#039;t know why this is happening, but please help me find the error in this approach.

lovagwb · Accepted Answer

Hint:   2sec⁡x−3csc⁡x=0 may not give us the minimum value as sec⁡xcsc⁡x is not constant   Use  4sec2x+9csc2x=4(1+tan2x)+9(1+cot2x)=13+(2tan⁡x−3cot⁡x)2+2⋅2⋅3

Vasquez · Accepted Answer

Let the minimum value be a. Then: 4sec2⁡x+9csc2⁡x=a⇒4sin2⁡x+9cos2⁡x=acos2⁡xsin2⁡x⇒4−4cos2⁡x+9cos2⁡x=acos2⁡x(1−cos2⁡x)⇒−acos4⁡x+(a−5)cos2⁡x−4=0 and since △=0,(a−5)2−4(−a)(−4)=0 or a2−26a+25=0⇒a=1,25. But since 4sec2⁡x+9csc2⁡x≥9csc2⁡x=9sin2⁡x≥91, the minimum value must be 25. To check if the minimum value is attained in the first quadrant, substitute a=25 into the quadratic equation above, which must result in a perfect square as △=0

user_27qwe · Accepted Answer

substitute cos2⁡(x)=t Having sec⁡(x)=1cos⁡(x)cosec(x)=1sin⁡(x)4cos2⁡x+9sin2⁡x→min, 0&amp;lt;x&amp;lt;π4sin2⁡xsin2⁡xcos2⁡x+9cos2⁡xsin2⁡xcos2⁡x→min substitute t=cos2⁡(x),sin2⁡(x)=1−t 4+5t(1−t)t→min finding minimum by differentiation ddt4+5t(1−t)t=0 5t2+8t−4t2(1−t)2=0 hence 5t2+8t−4=0, t≠0, t≠1 solving we get t=25, −2 Skipping verification that this value is minimum, we get cos2⁡x=25x=arccos⁡(25) and minimum value 4+5t(1−t)t=635×25=25

Minimizing 4\sec^2(x)+9\csc^2(x) for x in the first qu

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