Cot(1358)+tan(3608)=?
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1
Answer:
cot(−1358
0
)+tan(3608
0
)=−cot(8(180
0
)+82)+tan(20(180)+8) (since, cot(−θ)=−cot(θ))
=−cot(2(4)(180
0
)+82)+tan(2(10)(180)+8)
=−cot(82
0
)+tan(8
0
) (since, cot(2nπ+θ)=cot(θ) and tan(2nπ+θ)=tan(θ))
=−cot(90
0
−8
0
)+tan(8
0
)
=−tan(8
0
)+tan(8
0
) (since, cot(
2
π
−θ)=tan(θ)
=0
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