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So I have this rather simple trigonometric identity that, for the life of me, I cannot solve. I have worked on it for about 2 hours and still can't get it.

Show that $$\frac {2 - \csc^2 A}{\csc^2 A\space + \space2\cot A} \equiv \frac {\sin A \space -\space \cos A} {\sin A \space+\space \cos A}$$

Here's what I've done so far:

\begin{align} {2-\csc^2 A \over \csc^2 A+2\cot A} & = {2 - ({1 \over \sin A })^2 \over ({1 \over \sin A })^2 + 2({1 \over \tan A})} \\ &= {{2\sin^2 A \over \sin^2 A} - {1\over \sin^2 A} \over {1\over \sin^2 A} + 2({\cos A \over \sin A})} \\ & = {{2\sin^2 A -1 \over \sin^2 A} \over {1\over sin^2 A}+2({\sin A \cos A \over \sin^2 A})}\\ & = {2 \sin ^2 A -1 \over \sin^2 A} \times {\sin^2 A \over 2\sin A \cos A +1} \\ & = {2\sin^2 A -1 \over 2\sin A \cos A + 1} \end{align}

Lots of fractions are involved so I fear I may have made a mistake somewhere.


If anyone has any tips on proving these trigonometric identities, could they please add them in their answer? I've been told just to keep trying; though I believe there must be some 'troubleshooting' method to finish the problem.

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  • $\begingroup$ Well, the good news is that your manipulations so far are valid. $\endgroup$ – Daniel Martin Oct 16 '17 at 10:19
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You are nearly there! Let us go on with the identity $\sin^2 + \cos^2 = 1$ and some clever factorizations:

$$\frac{2 \sin^2 A - 1}{2 \sin A \cos A + 1} = \frac{\sin^2 A - \cos^2 A}{2\sin A \cos A + \sin^2 A + \cos^2 A} = \frac{(\sin A + \cos A)(\sin A - \cos A)}{(\sin A + \cos A)^2} = \frac{\sin A - \cos A}{\sin A + \cos A}. $$

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Good till now.

Now write

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

so denominator becomes whole square of form ${ (a+b)^2}$ and numerator is of form ${a^2-b^2}$

Cancel out (sinA + cosA)

and you are good to go.

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=$\frac{2 sin^2(A)-1}{1+2sin(A)cos(A)}$

=$\frac{2sin^2(A)-(cos^2(A)+sin^2(A))}{(cos^2(A)+sin^2(A))+2sin(A)cos(A)}$

=$\frac{sin^2(A)-cos^2(A)}{(sin(A)+cos(A))^2}$

=$\frac{(sin(A)+cos(A))(cos(A)+sin(A))}{(sin(A)+cos(A))^2}$

=$\frac{sin(A)-cos(A)}{(sin(A)+cos(A))}$

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