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Question

If X=sin(θ+7π12)+sin(θπ12)+sin(θ+3π12),Y=cos(θ+7π12)+cos(θπ12)+cos(θ+3π12), then prove that XYYX=2tan2θ.

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Solution

Given that,

X=sin(θ+7π12)+sin(θπ12)+sin(θ+3π12)

Y=cos(θ+7π12)+cos(θπ12)+cos(θ+3π12)

Then,

X=sin(θ+7π12)+sin(θπ12)+sin(θ+3π12)

=2sin(θ+7π12+θπ12)2cos(θ+7π12θ+π12)2+sin(θ+3π12)

=2sin(2θ+6π12)2cos8π24+sin(θ+3π12)

=2sin(θ+π4)cosπ3+sin(θ+3π12)

=2×12sin(θ+π4)+sin(θ+3π12)

=sin(θ+π4)+sin(θ+3π12)

X=2sin(θ+π4)

Similarly,

Y=cos(θ+7π12)+cos(θπ12)+cos(θ+3π12)

=2cos(θ+7π12+θπ12)2cos(θ+7π12θ+π12)2+cos(θ+3π12)

=2cos(θ+π4)cosπ3+cos(θ+3π12)

=cos(θ+π4)+cos(θ+π4)

Y=2cos(θ+π4)

Put the value of X&Y in LHS we get,

LHS

XYYX

=2sin(θ+π4)2cos(θ+π4)2cos(θ+π4)2sin(θ+π4)

=sin(θ+π4)cos(θ+π4)cos(θ+π4)sin(θ+π4)

=tan(θ+π4)cot(θ+π4)

=tanθ+tanπ41tanθtanπ4coθcotπ41coθ+cotπ4

=1+tanθ1tanθcotθ1cotθ+1

=1+tanθ1tanθ1tanθ11tanθ+1

=1+tanθ1tanθ1tanθ1+tanθ

=(1+tanθ)2(1tan2θ)1tan2θ

=1+tan2θ+2tanθ1tan2θ+2tanθ1tan2θ

=4tanθ1tan2θ

=2×2tanθ1tan2θ

=2tan2θ

=RHS

Hence, proved.

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