UPSC CDS 1 2018 Math Paper

x=

sec2θ−tan‌θ

sec2θ+tan‌θ

=

1−sin‌θ‌cos‌θ

1+sin‌θ‌cos‌θ

Now,
−

1

2

≤sin‌θ‌cos‌θ≤

1

2

(As −1≤sin‌2‌θ≤1)
∴ x will be minimum when sin‌θ‌cos‌θ=

1

2

⇒ Minimum value of x=

(1− 1 2 )

(1+ 1 2 )

=

( 1 2 )

( 3 2 )

=

1

3

And x will be maximum when sin‌θ‌cos‌θ=

−1

2

⇒ Maximum value of x=

(1+ 1 2 )

(1− 1 2 )

=

( 3 2 )

( 1 2 )

=3
∴

1

3

≤x≤3
Trick :
Take θ =45°.Then
x=

sec245°−tan‌45°

sec245°+tan‌45°

=

2−1

2+1

=

1

3

Clearly, options (a), (b), (c) are not correct.
∴ Option (d) is correct.