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cot(a+b)= cotacotb-1 (over) cota+cotb?

cot(a+b)= cotacotb-1 (over) cota+cotb?

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2 Answers

Meghna Shrivastava
18 Points
10 years ago

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
askIITians Faculty 629 Points
one year ago
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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