Question

$$\left. \begin{array} { l } { \tan 35 ^ { \circ } = } \\ { k } \\ { \tan ( 45 - \tan ) } \\ { 1 + \tan 145 } \\ { \tan 125 ^ { \circ } } \end{array} \right.$$

Answer

$$a=nSdeg/(15625*145_ta*n^3*t)$$

Solution


Remove parentheses.
\[tanSdeg=tan\times 145_tan\times 125tan\times 125\]
Cancel \(ta\) on both sides.
\[nSdeg=n\times 145_tan\times 125tan\times 125\]
Take out the constants.
\[nSdeg=(125\times 125)nnnta\times 145_ta\]
Simplify  \(125\times 125\)  to  \(15625\).
\[nSdeg=15625nnnta\times 145_ta\]
Simplify  \(15625nnnta\times 145_ta\)  to  \(15625{n}^{3}ta\times 145_ta\).
\[nSdeg=15625{n}^{3}ta\times 145_ta\]
Regroup terms.
\[nSdeg=15625\times 145_ta{n}^{3}ta\]
Divide both sides by \(15625\).
\[\frac{nSdeg}{15625}=145_ta{n}^{3}ta\]
Divide both sides by \(145_ta\).
\[\frac{\frac{nSdeg}{15625}}{145_ta}={n}^{3}ta\]
Simplify  \(\frac{\frac{nSdeg}{15625}}{145_ta}\)  to  \(\frac{nSdeg}{15625\times 145_ta}\).
\[\frac{nSdeg}{15625\times 145_ta}={n}^{3}ta\]
Divide both sides by \({n}^{3}\).
\[\frac{\frac{nSdeg}{15625\times 145_ta}}{{n}^{3}}=ta\]
Simplify  \(\frac{\frac{nSdeg}{15625\times 145_ta}}{{n}^{3}}\)  to  \(\frac{nSdeg}{15625\times 145_ta{n}^{3}}\).
\[\frac{nSdeg}{15625\times 145_ta{n}^{3}}=ta\]
Divide both sides by \(t\).
\[\frac{\frac{nSdeg}{15625\times 145_ta{n}^{3}}}{t}=a\]
Simplify  \(\frac{\frac{nSdeg}{15625\times 145_ta{n}^{3}}}{t}\)  to  \(\frac{nSdeg}{15625\times 145_ta{n}^{3}t}\).
\[\frac{nSdeg}{15625\times 145_ta{n}^{3}t}=a\]
Switch sides.
\[a=\frac{nSdeg}{15625\times 145_ta{n}^{3}t}\]