jamers Posted October 17, 2011 Share Posted October 17, 2011 hello i am in grade 12 advanced functions and i help with this trigonomatrie question. prove: Sec^2(teta) + cosec^2(teta) = cosec^2(teta) x sec^2(teta) for the first step i picked the left side to prove, 1/cos(teta)x 1/cos(teta) + 1/sin(teta) x 1/sin(teta) is this right? could somone help me wit hthe next step thanks. Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted October 17, 2011 Share Posted October 17, 2011 Yes, I think it's true that: [math]\sec^2 \theta + \csc^2 \theta = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}[/math] You can do the same thing with the other side, turning secant and cosecant into sines and cosines. From there, I think you should just try adding the fractions together. Link to comment Share on other sites More sharing options...

csmyth3025 Posted October 18, 2011 Share Posted October 18, 2011 (edited) Hello. I am in grade 12 advanced functions and I [need] help with this trigonomatrie question. prove: Sec^2(theta) + cosec^2(theta) = cosec^2(theta) x sec^2(theta) for the first step I picked the left side to prove, 1/cos(theta)x 1/cos(theta) + 1/sin(theta) x 1/sin(theta) is this right? could someone help me with the next step thanks. High school trigonometry is about the limit of my math knowledge, so please bear with me if my explanation seems overly simplified. Like you, to find an answer I have to "logic it out". The fundamental trigonometric relationships that apply to your problem as originally stated are these: Secant= hypotenuse/side adjacent (sec=c/a) -and- Cosecant=hypotenuse/side opposite (csc=c/b) Where a=side adjacent, b=side opposite, and c=hypotenuse. As normally noted in advanced trigonometry a=x axis, b=y axis, c=r (the length of the line from the origin to the x,y co-ordinates), and theta= the central angle. As a practical example, there is a simple type of right triangle with a 3-4-5 relationship of sides, where we can say that a (the side adjacent)=3, b (the side opposite)=4, and c (the hypotenuse)=5. Putting these numbers in the original equation gives you: (c/a)^{2} + (c/b)^{2} = (c/b)^{2} x (c/a)^{2 }= 25/9 + 25/16 = 25/16 x 25/9, cross-multiplying the left side and multiplying through on the right side gives: 625/144 = 625/144 This example certainly shows the equivalence that the question poses for this particular triangle. The trick is to show algebraically that in all cases (c/a)^{2} + (c/b)^{2} = (c/b)^{2} x (c/a)^{2}. Inverting these relationships, as you've done, isn't necessary to the solution. Hint: a^{2} + b^{2} = c^{2} ^{ } Chris PS The logic expected by your instructor is probably more sophisticated than I've presented, but it's the best I can do. Edited to add PS Edited October 18, 2011 by csmyth3025 Link to comment Share on other sites More sharing options...

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