Question

Simplify sin Θ/√1-sin^2 Θ

Answer 1

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Simplify the trigonometric expression.

Call the expression E:

$\mathsf{E=\dfrac{sin\,\theta}{\sqrt{1-sin^2\,\theta}}\qquad\qquad(but~~1-sin^2\,\theta=cos^2\,\theta)}\\\\\\ \mathsf{E=\dfrac{sin\,\theta}{\sqrt{cos^2\,\theta}}}\\\\\\ \mathsf{E=\dfrac{sin\,\theta}{\left|cos\,\theta\right|}}}$


$\mathsf{E}=\left\{\!\begin{array}{rl}\mathsf{\dfrac{sin\,\theta}{cos\,\theta}\,,}&\mathsf{\quad if~~cos\,\theta>0}\\\\\mathsf{\dfrac{sin\,\theta}{-\,cos\,\theta}\,,}&\mathsf{\quad if~~cos\,\theta$


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

Answer 2

Answer:

$\tan\theta$

Step-by-step explanation:

We are given our expression as

$\frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}$

Here we have to recall the trignometric identities which says

$\sin^{2}\theta + \cos^{2}\theta=1$

taking $\sin^{2}\theta$ to the right hand side of =

$\cos^{2}\theta = 1- \sin^{2}\theta$

Hence we replace $1- \sin^{2}\theta$ with $\cos^{2}\theta$ in our main equation,

$\frac{\sin\theta}{\sqrt{\cos^{2}\theta}}\\\frac{\sin\theta}{\cos\theta}\\=\tan\theta$

Hence the answer is $\tan\theta$