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Simplify the trigonometric expression.
Call the expression E:
\(\mathsf{E=\dfrac{sin\,\theta}{\sqrt{1-sin^2\,\theta}}\qquad\qquad(but~~1-sin^2\,\theta=cos^2\,\theta)}\\\\\\
\mathsf{E=\dfrac{sin\,\theta}{\sqrt{cos^2\,\theta}}}\\\\\\
\mathsf{E=\dfrac{sin\,\theta}{\left|cos\,\theta\right|}}}\)
\(\mathsf{E}=\left\{\!\begin{array}{rl}\mathsf{\dfrac{sin\,\theta}{cos\,\theta}\,,}&\mathsf{\quad if~~cos\,\theta>0}\\\\\mathsf{\dfrac{sin\,\theta}{-\,cos\,\theta}\,,}&\mathsf{\quad if~~cos\,\theta<0}\end{array}\right.\\\\\\\\\mathsf{E}=\left\{\!\begin{array}{rr}\mathsf{tan\,\theta,\qquad if}&\mathsf{-\dfrac{\pi}{2}+k\cdot 2\pi<\theta<\dfrac{\pi}{2}+k\cdot 2\pi}\\\\\mathsf{-\,tan\,\theta,\qquad if}&\mathsf{\dfrac{\pi}{2}+k\cdot 2\pi<\theta<\dfrac{3\pi}{2}+k\cdot 2\pi}\end{array}\right.\qquad\mathsf{k \in \mathbb{Z}}\)
So basically E is equivalent to \(\pm\,\mathsf{tan\,\theta},\) where the sign depends on which quadrant \(\theta\) lies. Shortly,
\(\mathsf{\dfrac{sin\,\theta}{\sqrt{1-sin^2\,\theta}}}=\left\{\!
\begin{array}{rl}
\mathsf{tan\,\theta,}&\mathsf{\quad if~\theta~lies~either~in~the~1^{st}~or~the~4^{th}~quadrant}\\\\
\mathsf{-\,tan\,\theta,}&\mathsf{\quad if~\theta~lies~either~in~the~2^{nd}~or~the~3^{rd}~quadrant}
\end{array}\)
I hope this helps. =)
Tags: simplify trigonometric trig expression sine cosine tangent sin cos tan absolute value modulus trigonometry
