Question : If $0°<A<90°$, the value of $\frac{\tan A\ -\ \sec A\ -\ 1}{\tan A\ +\ \sec A\ +\ 1}$ is:
Option 1: $\frac{\sin A-1}{\cos A}$
Option 2: $\frac{1-\sin A}{\cos A}$
Option 3: $\frac{1-\cos A}{\sin A}$
Option 4: $\frac{\sin A+1}{\cos A}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
Correct Answer: $\frac{\sin A-1}{\cos A}$
Solution :
Given: $\frac{\tan A-\sec A-1}{\tan A+\sec A+1}$
= $\frac{\tan A-\sec A-(\sec^{2} A-\tan^{2} A)}{\tan A+\sec A+1}$
= $\frac{(\tan A-\sec A)-(\sec A-\tan A)(\sec A+\tan A)}{\tan A+\sec A+1}$
= $\frac{(\tan A-\sec A)+(\tan A-\sec A)(\sec A+\tan A)}{\tan A+\sec A+1}$
= $\frac{(\tan A-\sec A)(1+\sec A+\tan A)}{\tan A+\sec A+1}$
= $\tan A-\sec A$
= $\frac{\sin A}{\cos A}-\frac{1}{\cos A}$
= $\frac{\sin A-1}{\cos A}$
Hence, the correct answer is $\frac{\sin A-1}{\cos A}$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
