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Question

Prove that: cot712=2+3+4+6

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Solution

cot712=2+3+4+6
L.H.S=cot712=cot152
=cos(152)sin(152)=2cos(152)2sin(152)
=2cos2(152)2sin(152)cos(152)=1+cos15°sin15°
[1+cos2θ=2cos22θ2]
[&2sinθ2.cosθ2=sin2θ2]
cos15°=14(6+2)&sin15°=14(62)
LHS=1+14(6+2)14(62)=4+6+262
Apply componendo and dividendo, we get
LHS=(4+6+2)(6+2)(62)(6+2)
=46+42+6+12+12+262
=46+42+8+23+234
=84+424+434+464
=2+2+3+6
=2+3+4+6
R.H.S
Hence, the answer is prove.

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