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What is (\( \Large \frac{sin x-cos x+1}{sin x + cos x -1} \) ) eqaul to?


A) (\( \Large \frac{sin x-1}{ cos x} \) )

B) (\( \Large \frac{sin x+1}{ cos x} \) )

C) (\( \Large \frac{sin x-1}{ cos x+1} \) )

D) (\( \Large \frac{sin x+1}{ cos x+1} \) )

Correct Answer:
B) (\( \Large \frac{sin x+1}{ cos x} \) )

Description for Correct answer:
\( \Large \frac{sinx - cosx + 1}{sinx + cosx - 1} \times \frac{sinx + cosx + 1}{sinx + cosx + 1} \)

\( \Large = \frac{[(sinx + 1) - cosx] \times [(sinx + 1)+cosx]}{(sinx + cosx)^{2} - 1} \)

=\( \Large \frac{(sinx + 1)^{2} - cos^{2}x}{2sinxcosx} \)

=\( \Large \frac{sin^{2}x + 1 + 2sinx - cons^{2}x }{2sinxcosx} \)

=\( \Large \frac{2sin^{2}x + 2sinx }{2sinxcosx} \)

=\( \Large \frac{2sinx(sinx + 1 )}{2sinxcosx} \)

=\( \Large \frac{sinx + 1}{cosx} \)

option b is correct

Part of solved CDS Maths(1) questions and answers : Exams >> CDSE >> CDS Maths(1)




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