Example

3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b

Question

$$\left. \begin{array} { l } { \cot ( A + 45 ^ { \circ } ) - \tan ( A - 45 ^ { \circ } ) = \frac { 2 \cos 2 A } { 1 + \sin 2 A } } \end{array} \right.$$

Answer

$$a=((2*cos(2)*A)/(1+sin(2)*A)+tan(A-45*deg))/(t^2*h*cot(A+45*deg))$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{t}^{2}ha\cot{(A+45deg)}-\tan{(A-45deg)}=\frac{2\cos{2}A}{1+\sin{2}A}\]
Add \(\tan{(A-45deg)}\) to both sides.
\[{t}^{2}ha\cot{(A+45deg)}=\frac{2\cos{2}A}{1+\sin{2}A}+\tan{(A-45deg)}\]
Divide both sides by \({t}^{2}\).
\[ha\cot{(A+45deg)}=\frac{\frac{2\cos{2}A}{1+\sin{2}A}+\tan{(A-45deg)}}{{t}^{2}}\]
Divide both sides by \(h\).
\[a\cot{(A+45deg)}=\frac{\frac{\frac{2\cos{2}A}{1+\sin{2}A}+\tan{(A-45deg)}}{{t}^{2}}}{h}\]
Simplify  \(\frac{\frac{\frac{2\cos{2}A}{1+\sin{2}A}+\tan{(A-45deg)}}{{t}^{2}}}{h}\)  to  \(\frac{\frac{2\cos{2}A}{1+\sin{2}A}+\tan{(A-45deg)}}{{t}^{2}h}\).
\[a\cot{(A+45deg)}=\frac{\frac{2\cos{2}A}{1+\sin{2}A}+\tan{(A-45deg)}}{{t}^{2}h}\]
Divide both sides by \(\cot{(A+45deg)}\).
\[a=\frac{\frac{\frac{2\cos{2}A}{1+\sin{2}A}+\tan{(A-45deg)}}{{t}^{2}h}}{\cot{(A+45deg)}}\]
Simplify  \(\frac{\frac{\frac{2\cos{2}A}{1+\sin{2}A}+\tan{(A-45deg)}}{{t}^{2}h}}{\cot{(A+45deg)}}\)  to  \(\frac{\frac{2\cos{2}A}{1+\sin{2}A}+\tan{(A-45deg)}}{{t}^{2}h\cot{(A+45deg)}}\).
\[a=\frac{\frac{2\cos{2}A}{1+\sin{2}A}+\tan{(A-45deg)}}{{t}^{2}h\cot{(A+45deg)}}\]
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