if sin A=4/5 and sin B=5/13 then cot(A-B)=?
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Answered by
6
sinA=4/5 , cotA= 3/4
sinB=5/13 , cotB=12/5
cot(A-B)=( cotA.cotB + 1)/(cotB - cotA)
=( 3/4×12/5 + 1)/(12/5 - 3/4)
=(9/5 + 1)/(48-15/20)
=(9+5)5/(33/20)
=14×4/33
=56/33
sinB=5/13 , cotB=12/5
cot(A-B)=( cotA.cotB + 1)/(cotB - cotA)
=( 3/4×12/5 + 1)/(12/5 - 3/4)
=(9/5 + 1)/(48-15/20)
=(9+5)5/(33/20)
=14×4/33
=56/33
veeracharan9:
ans is 56/33
sorry okay
try it
ho gya dekh lo
please try
yrr ho gya dekh lo correct kr diya
ok thanks
exalent bro
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Answered by
0
Answer:
Step-by-step explanation:
sinA=4/5 , cotA= 3/4
sinB=5/13 , cotB=12/5
cot(A-B)=( cotA.cotB + 1)/(cotB - cotA)
=( 3/4×12/5 + 1)/(12/5 - 3/4)
=(9/5 + 1)/(48-15/20)
=(9+5)5/(33/20)
=14×4/33
=56/33
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