Prove that:
$$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1 + \sin A}{ \cos A}$$
I found this difficult for some reason. I tried subsituting tan A for sinA / cos A and sec A as 1/cos A and then simplifying but it didn't work.
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$\begingroup$ Sorry, I had a tough time with the MathJAX $\endgroup$ – user2130295 Jan 3 '14 at 12:19
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$\begingroup$ You are doing well with the MathJAX now. Putting a backslash before the functions will render them in the proper font. So \cos gives $\cos$ instead of cos giving $cos$ $\endgroup$ – Ross Millikan Jan 3 '14 at 12:20
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$\begingroup$ Try to simplify the left hand side. $\endgroup$ – mathlove Jan 3 '14 at 12:20
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Setting $\displaystyle1=\sec^2A-\tan^2A$ in the numerator,
So, the numerator becomes $$\tan A+\sec A-1$$ $$=\tan A+\sec A-(\sec^2A-\tan^2A)$$ $$=(\sec A+\tan A)\{1-(\sec A-\tan A)\}$$ $$=\frac{(1+\sin A)}{\cos A}(1-\sec A+\tan A)$$
answered Jan 3 '14 at 12:20
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$\begingroup$ @user2130295, my pleasure. Related : math.stackexchange.com/questions/426981/… $\endgroup$ – lab bhattacharjee Jan 3 '14 at 12:37
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You may also simplify the expression by using the definition of the tangent and the secant function:
\begin{align} &\dfrac{\tan A +\sec A-1}{\tan A - \sec A +1} \\=&\dfrac{\dfrac{\sin A}{\cos A} +\dfrac1{\cos A}-1}{\dfrac{\sin A}{\cos A} - \dfrac1{\cos A} +1} \\=&\dfrac{\sin A -\cos A+1}{\sin A+\cos A-1} \\=&\dfrac{(\sin A -\cos A+1)\cdot(\sin A+\cos A+1)}{(\sin A+\cos A-1)\cdot(\sin A+\cos A+1)}=\dfrac{(\sin A +1)^2-\cos^2 A}{(\sin A+\cos A)^2-1} \\=&\dfrac{\sin^2 A+(1-\cos^2 A) +2\sin A}{2\sin A\cos A}\end{align}
The rest is fairly straightforward.
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Multiplying the numerator & the denominator by $\cos A$
$$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}$$
$$\cos A(\sin A+1-\cos A)=\cos A\sin A+\cos A-\cos^2A$$ $$=\cos A(1+\sin A)-(1-\sin^2A)$$ $$\implies\cos A(\sin A+1-\cos A)=(1+\sin A)(\cos A-1+\sin A)$$
$$\implies\frac{\sin A+1-\cos A}{\sin A-1+\cos A}=?$$
We can start with $(1+\sin A)(\sin A-1+\cos A)$ as well
Find some similar results here
answered Jan 3 '14 at 19:11
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$\begingroup$ @user2130295, here is another method $\endgroup$ – lab bhattacharjee Jan 3 '14 at 19:13
