answer for 1st question pls asap
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Aksha02:
pls fast
Ar(abc)/ar( qrp)=bc
Bc^2/pr^2
Let theta A
6CotA+2CosecA=cotA+5cosecA. 6CotA -cotA=5CosecA -2cosec. 5CotA=3Cosec. 5CosA/SinA= 3(1/sinA). cosA/sinA*sinA/1= 3/5. from sinA sinA will cancel and value of cos=3/5
Here your steps for ques.4
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Ans of que. 1
The remainder is 3
The remainder is 3
but i need the steps
can u pls do tje the 6 th too?
Its ans is 10 que 6
Can u tell me 4 question
Ques no 4 ans is 3/5
yes crct for ans 4
can u pls do with steps n post
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