Grade 10Trigonometry
prove the following cot2A.cot2B+cot2B.cot2C+cot2C.cot2A=1
prove the following cot2A.cot2B+cot2B.cot2C+cot2C.cot2A=1
1 Answer
Arun
Dear Aryan
a + b + c = π; so (a + b) = π - c
==> cot(a + b) = cot(π - c) = -cot(c)
==> {cot(a)*cot(b) - 1}/{cot(a) + cot(b)} = -cot(c)
Simplifying the above,
==> cot(a + b) = cot(π - c) = -cot(c)
==> {cot(a)*cot(b) - 1}/{cot(a) + cot(b)} = -cot(c)
Simplifying the above,
cot(a)*cot(b) + cot(b)*cot(c) + cot(c)*cot(a) = 1
Regards
Arun(askIITians forum expert)
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