SSC CGL Tier-1 16-08-2015 Shift 2 Solved Paper

sec θ + tan θ = 2 + 5
Since sec2 θ – tan2 θ = 1
⇒ (sec θ + tan θ ) (sec θ – tan θ ) = 1
⇒ sec θ – tan θ =

1

√5+2

=

1

√5+2

×

√5−2

√5−2

=

√5−2

5−4

= √5 - 2
∴ sec θ + tan θ + sec θ - tan θ = 2 + √5 + √5 - 2
⇒ 2 sec θ = 2√5
⇒ sec θ = √5 ... (i)

sec θ + tan θ - (sec θ - tan θ)
= 2 + √5−√5 + 2
⇒ 2 tan θ = 4
⇒ tan θ = 2 ... (ii)
∴ sin θ =

tanθ

secθ

=

2

√5