UPSC NDA Math Model Paper 6

Consider the relation in the set P={θ:sin‌θ−cos‌θ=√2‌cos‌θ}
sin‌θ−cos‌θ=√2‌cos‌θ
Dividing both sides by cos θ, we get:
tan‌θ−1=√2
⇒tan‌θ=√2+1
∴P={θ:tan‌θ=√2+1} .......(1)
Consider the relation in the set
Q={θ:sin‌θ+cos‌θ=√2‌sin‌θ}.
sin‌θ+cos‌θ=√2‌sin‌θ
Dividing both sides by cos‌θ, we get:
tan‌θ+1=√2‌tan‌θ
⇒tan‌θ=

1

√2−1

=

1

√2−1

×

√2+1

√2+1

=

√2+1

2−1

=√2+1
∴Q={θ:tan‌θ=√2+1} .......(2)
Comparing equations (1) and (2), we have:
P = Q.