JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 7)
Let P = {$$\theta $$ : sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 \,\cos \theta $$}
and Q = {$$\theta $$ : sin$$\theta $$ + cos$$\theta $$ = $$\sqrt 2 \,\sin \theta $$} be two sets. Then
and Q = {$$\theta $$ : sin$$\theta $$ + cos$$\theta $$ = $$\sqrt 2 \,\sin \theta $$} be two sets. Then
P $$ \subset $$ Q and Q $$-$$ P $$ \ne $$ $$\phi $$
Q $$ \not\subset $$ P
P $$ \not\subset $$ Q
P = Q
Explanation
Given,
sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 $$cos$$\theta $$
$$ \Rightarrow $$ sin$$\theta $$ = $$\left( {\sqrt 2 + 1} \right)$$cos$$\theta $$
$$ \Rightarrow $$ sin$$\theta $$ = $${{2 - 1} \over {\sqrt 2 - 1}}$$cos$$\theta $$
$$ \Rightarrow $$ $$\left( {\sqrt 2 - 1} \right)$$sin$$\theta $$ = cos$$\theta $$ . . . (1)
This is for set P.
sin$$\theta $$ + cos$$\theta $$ = $${\sqrt 2 }$$sin$$\theta $$
$$ \Rightarrow $$ cos$$\theta $$ = $$\left( {\sqrt 2 - 1} \right)$$sin$$\theta $$ . . . .(2)
This is from set Q.
So, P = Q
sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 $$cos$$\theta $$
$$ \Rightarrow $$ sin$$\theta $$ = $$\left( {\sqrt 2 + 1} \right)$$cos$$\theta $$
$$ \Rightarrow $$ sin$$\theta $$ = $${{2 - 1} \over {\sqrt 2 - 1}}$$cos$$\theta $$
$$ \Rightarrow $$ $$\left( {\sqrt 2 - 1} \right)$$sin$$\theta $$ = cos$$\theta $$ . . . (1)
This is for set P.
sin$$\theta $$ + cos$$\theta $$ = $${\sqrt 2 }$$sin$$\theta $$
$$ \Rightarrow $$ cos$$\theta $$ = $$\left( {\sqrt 2 - 1} \right)$$sin$$\theta $$ . . . .(2)
This is from set Q.
So, P = Q
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