UPSC 2015 CDS I Math Paper

1

1−sin‌x

=4+2√3
⇒1−sin‌x=

1

4+2√3

⇒sin‌x=1−

1

4+2√3

⇒sin‌x=

3+2√3

4+2√3

⇒sinx=

√3(√3+2)

2+2√3

⇒sin‌x=

√3

2

⇒x=

π

3

Hence, statement (1) is incorrect. We know that −1≤sin‌x≤1
Also, 0≤sin2x≤1⇒1≤3sin2x≤3
The given equation holds good if sin‌x=1 and 3sin2x=1.But, there exists no value of x for which sin‌x=1 and 3sin2x=1 simultaneously.
Thus, the given equation does not hold good for any real x.
Hence, statement (2) is correct