Trigonometry
cot(a+b)= cotacotb-1 (over) cota+cotb?
cot(a+b)= cotacotb-1 (over) cota+cotb?
2 Answers
Meghna Shrivastava
Taking L.H.S.,
cot(a+b)=1/tan(a+b)
=1-tan(a)+tan(b)/tan(a)+tan(b)
[Since,tan(a+b)=tan(a)+tan(b)/1-tan(a)tan(b)]
Dividing by tan(a)tan(b) in num and deno,we have
=cot(a)cot(b)-1/cot(a)+cot(b)
=R.H.S.
Hence Proved.
Kushagra Madhukar
Dear student,
Please find the attached solution to your question.
L.H.S = cot(a+b)=1/tan(a+b)
=1 – tan(a) + tan(b) / tan(a) + tan(b)
[ tan(a+b) = tan(a) + tan(b) / 1 – tan(a)tan(b)]
Dividing by tan(a)tan(b) in num and deno,we have
= cot(a)cot(b) – 1 / cot(a) + cot(b)
= R.H.S.
Hope it helps.
Thanks and regards,
Kushagra
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