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Prove that $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$


$$\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{\frac{2}{\cos\theta} +\frac{3\sin\theta}{\cos\theta}+5\sin\theta-7\cos\theta+5}{\frac{2\sin\theta}{\cos\theta} +\frac{3}{\cos\theta}+5\cos\theta+7\sin\theta+8}$$

$$=\frac{2+3\sin\theta+5\sin\theta\cos\theta-7\cos^2\theta+5\cos\theta}{2\sin\theta+3+5\cos^2\theta+7\sin\theta\cos\theta+8\cos\theta}$$

I am stuck here.I tried to factorize numerator and denominator but does not succeed.

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Let $\cos\theta=c,\sin\theta=s$. Then, $$\begin{align}&s(2+3s+5sc-7c^2+5c)-(1-c)(2s+3+5c^2+7cs+8c)\\&=5c^3+5cs^2-5c+3c^2+3s^2-3\\&=5c(c^2+s^2-1)+3(c^2+s^2-1)\\&=0\end{align}$$ from which the claim follows.

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Let's simplify our notation by writing $c=\cos\theta$ and $s=\sin\theta$. First note that $$\frac{1-\cos\theta}{\sin\theta}=\frac{\sin\theta}{1+\cos\theta}.$$ So, for arbitrary $a$ and $b$, not both zero, we can write either fraction as $$\frac{a(1-c)+bs}{as+b(1+c)}.$$It's not hard to spot that setting $a=5c$ and $b=3+5c+2s$ gives us the original expression.

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It is possible using the Rule $ P = n p, Q = n q $, $n$ any real number

and recognizing

$$ \frac{1-\cos\theta}{\sin\theta}= \frac{\sin\theta}{1+\cos\theta}\tag{0} ,$$

$$ \frac {u}{v} = \frac {p}{q} = \frac {P}{Q}=\frac {u \pm P}{v \pm Q}, $$ and entirely avoid all trigonometric calculations.

Multiply numerator and denominator by 7 and add

$$\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}=\frac{7-7 \cos\theta}{7 \sin\theta}\tag{1}$$

$$ F = \frac{2\sec\theta +3\tan\theta+5\sin\theta-2}{2\tan\theta +3\sec\theta+5\cos\theta+8} \tag{2} $$

Multiply numerator and denominator by 5 and subtract

$$\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{\sin\theta}{1+\cos\theta}=\frac{5 \sin\theta}{5+5\cos\theta} \tag{3} $$

$$ F=\frac{2\sec\theta +3\tan\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta +7\sin\theta+3} \tag{4}$$

Repeating the Rule by subtracting on (2) and (4)

$$ F=\frac{5 \sin \theta + 7\cos\theta-7}{5\cos\theta -7\sin\theta +5}\tag{5} $$

Repeating the rule by subtracting (2) from (3)

$$ F =\frac{5 \sin \theta + 7\cos\theta-7}{5\cos\theta -7\sin\theta +5}\tag{6} $$

which are identical.

It will be appreciated that it is simple algebra and there is no trig at all after (0) and probably that is how the problem had been set.

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hint $\frac{1-\cos(\theta)}{sin(\theta)}=\tan(\theta/2)$ so now convert everything to tan by using $\sin(2x)=\frac{2tan(x)}{1+\tan^2(x)},\cos(2x)=\frac{1-\tan^2(x)}{1+\tan^2(x)}$ then just simplify it. as $1+\tan^2(x)$ cancels off from the expression by taking lcm then use $tan(x)=t$ for algebraic simplifications. I dont think there's a nice short way to do it.

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  • $\begingroup$ I think you need a double angle on the left hand side of the two equations you give for sine and cosine. $\endgroup$ – Paul Sundheim Apr 12 '16 at 17:18
  • $\begingroup$ Yes thanks missed the $2$ $\endgroup$ – Archis Welankar Apr 12 '16 at 17:31

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