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Math Help - range question

  1. #1
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    range question

    range of
    sin x/(cosx)^2
    (i)without calculus
    (ii)with calculus
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  2. #2
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    Hello, ayushdadhwal!

    \text{Range of: }\,\dfrac{\sin x}{\cos^2x}

    \text{(i) without calculus}

    I substituted values: . x \:=\:0,\,\frac{\pi}{4},\,\frac{\pi}{2},\,\frac{3\p  i}{4},\,\pi,\,\frac{5\pi}{4},\,\hdots
    . . and found the following behavior:


    . . \begin{array}{|c|c|c|} \hline<br />
\text{Quadrant} & \sin x,\,\cos^2\!x & \dfrac{\sin x}{\cos^2x} \\ \hline \hline<br />
1 & \sin x\!: 0 \to 1 &  \\<br />
& \cos^2\!x\!: 1 \to 0 & 0 \to +\infty\\ \hline<br /> <br />
2 & \sin x\!: 1 \to 0 & \\<br />
&\cos^2\!x\!: 0 \to 1 & +\infty \to 0 \\ \hline<br /> <br />
3 & \sin x\!: 0 \to \text{-}1 & \\<br />
& \cos^2\!x\!: 1 \to 0 & 0 \to \text{-}\infty \\ \hline<br /> <br />
4 & \sin x\!: \text{-}1 \to 0 & \\<br />
& \cos^2\!x\!: 0 \to 1 & \text{-}\infty \to 0 \\ \hline <br />
\end{array}


    \text{Therefore, the range of }\,\dfrac{\sin x}{\cos^2\!x}\,\text{ is: }\,(\text{-}\infty,\,+\infty)




    \text}(ii) with calculus}

    \text{We note that: }\:\dfrac{\sin x}{\cos^2\!x} \:=\:\sec x\tan x


    \text{This is the }derivative\text{ of: }\:f(x) \,=\,\sec x

    \text{The graph of }\sec x\text{ looks like this:}\begin{array}{c} \cup\;\; \cup \\ \quad\;\cap\;\; \cap \end{array}

    \text{And its }slope\text{ ranges from }\text{-}\infty \text{ to }+\infty.

    \text{Therefore, the range of }\dfrac{\sin x}{\cos^2\!x}\text{ is: }\,(\text{-}\infty,\,+\infty)

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