Question

$$\frac { \sec 16 A - 1 } { \sec 8 A - 1 } = \frac { \tan 16 A } { \tan 4 A }$$

Answer

t=(tan(4)*COSsec(16*A)-1)/(8*tan(16)*A*(sec(8*A)-1))

Solution


Cancel \(A\).
\[\frac{COS(\sec{16A}-1)}{\sec{8A}-1}=\frac{\tan{16}}{\tan{4}}t\times 8A\]
Simplify  \(\frac{\tan{16}}{\tan{4}}t\times 8A\)  to  \(\frac{(\tan{16})t\times 8A}{\tan{4}}\).
\[\frac{COS(\sec{16A}-1)}{\sec{8A}-1}=\frac{(\tan{16})t\times 8A}{\tan{4}}\]
Regroup terms.
\[\frac{COS(\sec{16A}-1)}{\sec{8A}-1}=\frac{8(\tan{16})tA}{\tan{4}}\]
Multiply both sides by \(\tan{4}\).
\[\frac{COS(\sec{16A}-1)}{\sec{8A}-1}\tan{4}=8(\tan{16})tA\]
Use this rule: \(\frac{a}{b} \times c=\frac{ac}{b}\).
\[\frac{COS(\sec{16A}-1)\tan{4}}{\sec{8A}-1}=8(\tan{16})tA\]
Regroup terms.
\[\frac{\tan{4}COSsec(16A)-1}{sec(8A)-1}=8tan(16)tA\]
Divide both sides by \(8\).
\[\frac{\frac{\tan{4}COSsec(16A)-1}{sec(8A)-1}}{8}=tan(16)tA\]
Simplify  \(\frac{\frac{\tan{4}COSsec(16A)-1}{sec(8A)-1}}{8}\)  to  \(\frac{\tan{4}COSsec(16A)-1}{8(sec(8A)-1)}\).
\[\frac{\tan{4}COSsec(16A)-1}{8(sec(8A)-1)}=tan(16)tA\]
Divide both sides by \(\tan{16}\).
\[\frac{\frac{\tan{4}COSsec(16A)-1}{8(sec(8A)-1)}}{tan(16)}=tA\]
Simplify  \(\frac{\frac{\tan{4}COSsec(16A)-1}{8(sec(8A)-1)}}{tan(16)}\)  to  \(\frac{\tan{4}COSsec(16A)-1}{8tan(16)(sec(8A)-1)}\).
\[\frac{\tan{4}COSsec(16A)-1}{8tan(16)(sec(8A)-1)}=tA\]
Divide both sides by \(A\).
\[\frac{\frac{\tan{4}COSsec(16A)-1}{8tan(16)(sec(8A)-1)}}{A}=t\]
Simplify  \(\frac{\frac{\tan{4}COSsec(16A)-1}{8tan(16)(sec(8A)-1)}}{A}\)  to  \(\frac{\tan{4}COSsec(16A)-1}{8tan(16)A(sec(8A)-1)}\).
\[\frac{\tan{4}COSsec(16A)-1}{8tan(16)A(sec(8A)-1)}=t\]
Switch sides.
\[t=\frac{\tan{4}COSsec(16A)-1}{8tan(16)A(sec(8A)-1)}\]