A) \[\frac{1-\sin A}{\cos A}\]
B) \[\frac{1-\cos A}{\sin A}\]
C) \[\frac{1+\sin A}{\cos A}\]
D) \[\frac{1+\cos A}{\sin A}\]
Correct Answer: C
Solution :
\[\frac{\tan A+\sec A-1}{\tan A-\sec A+1}\] \[=\frac{\sin A-\cos A+1}{\sin A-1+\cos A}=\frac{\sin A+(1-\cos A)}{\sin A-(1-\cos A)}\] \[=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}+2{{\sin }^{2}}\frac{A}{2}}{2\sin \frac{A}{2}\cos \frac{A}{2}-2{{\sin }^{2}}\frac{A}{2}}\] \[=\frac{\cos \frac{A}{2}+\sin \frac{A}{2}}{\cos \frac{A}{2}-\sin \frac{A}{2}}=\frac{{{\left( \cos \frac{A}{2}+\sin \frac{A}{2} \right)}^{2}}}{{{\cos }^{2}}\frac{A}{2}-{{\sin }^{2}}\frac{A}{2}}\]\[=\frac{1+\sin A}{\cos A}\].
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