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$$
\text { Prove that } \frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A} \text {. }
$$
\text { Prove that } \frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A} \text {. }
$$
Solution:
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Verified Answer
We have LHS $=\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$ $=\frac{\tan A+\sec A-\left(\sec ^2 A-\tan ^2 A\right)}{(\tan A-\sec A+1)}$
$$
\begin{aligned}
&=\frac{\left(\because \sec ^2 A-\tan ^2 A=1\right]}{(1-\sec A+\tan A)} \\
&=\frac{(\sec A+\tan A)(1-\sec A+\tan A)}{1-\sec A+\tan A} \\
&=\sec A+\tan A=\frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}=\mathrm{RHS}
\end{aligned}
$$
$$
\begin{aligned}
&=\frac{\left(\because \sec ^2 A-\tan ^2 A=1\right]}{(1-\sec A+\tan A)} \\
&=\frac{(\sec A+\tan A)(1-\sec A+\tan A)}{1-\sec A+\tan A} \\
&=\sec A+\tan A=\frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}=\mathrm{RHS}
\end{aligned}
$$
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