AP ICET 2016 Paper

I Given : sec‌θ+cosec‌θ=5
⇒

1

cos‌θ

+

1

sin‌θ

=5
⇒sin‌θ+cos‌θ=5‌sin‌θ‌cos‌θ
⇒sin‌θ+cos‌θ=

5

2

sin2θ
Since sin 2θ or sinθ or cosθ is not given, we can’t determine sinθ + cosθ.
∴ I alone is not sufficient
II. Given : sin‌2‌θ=

1

2

⇒ 2θ = 30° or 150°[Sine is positive in QI and QII]
⇒ θ = 15° or 75°
sin‌15°+cos‌15°=sin‌15°+sin‌75°
Also, sin‌75°+cos‌75°=sin‌75°+sin‌15°
Hence, sin‌θ+cos‌θ can be uniquely determined.
∴ II alone is sufficient.
Aliter : sin‌θ+cos‌θ=√(sin‌θ+cosθ)2
√

3

2

we discard –

√3

2

as θ lies in the first quadrant.
∴ II alone is sufficient.