If A + B + C = 90°
Then,
tanA + tanB + tanA×tanB ×cotC is?
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Answers
Answered by
1
A+B+C = 90
A+B = 90-C
TAN(A+B) = TAN (90-C)
TANA + TANB / 1- TANATANB = COTC
TANA+ TANB = COTC - TANATANBCOTC
TANA + TANB + TANATANBCOTC = COTC
SO ANS = COTC
A+B = 90-C
TAN(A+B) = TAN (90-C)
TANA + TANB / 1- TANATANB = COTC
TANA+ TANB = COTC - TANATANBCOTC
TANA + TANB + TANATANBCOTC = COTC
SO ANS = COTC
HappiestWriter012:
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✌all the best :(
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Is it because you copied my answer or you're nagging me?
nagging?
what's the meaning
Please No Arguments :(
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Answered by
3
Hey there!
tanA + tanB + tanA×tanB ×cotC
= sinA /cosA +sinB /cosB +[ sin A SinB cos C]/cos A CosB SinC
=SinA CosB SinC +SinB CosA SinC +SinA SinB cosC / cos A CosB SinC
=SinA ( CosB SinC + SinB cos C) + SinB CosA SinC / cos A CosB SinC
=SinA Sin[B+C] + SinB CosA SinC / cos A CosB SinC
=SinA CosA + SinB CosA SinC / cos A CosB SinC
=CosA [ SinA + SinB SinC] / cos A CosB SinC
=CosA [ SinA + SinB SinC] / cos A CosB SinC
= SinA + SinB SinC / CosB SinC
Now, multiply and divide with 2
(2sinA+2sinBsinC)/2cosBsinC
=(2sinA+cos(B-C)-cos(B+C))/(sin(B+C)-sin(B-C))
=(2sinA+cos(B-C)-cos(90-A))/(sin(90-A)-sin(B-C))
=(2sinA+cos(B-C)-sinA)/(cosA-sin(B-C))
=(sinA+cos(B-C))/(CosA-sin(B-C))
=(sin(90-(B+C))+cos(B-C))/cos(90-(B+C))-sin(B-C))
=(cos(B+c)+cos(B-C))/(sin(B+C)-sin(B-C))
=2cosBcosC/2cosBsinC
=cosC/sinC
= CotC
hope helped!
tanA + tanB + tanA×tanB ×cotC
= sinA /cosA +sinB /cosB +[ sin A SinB cos C]/cos A CosB SinC
=SinA CosB SinC +SinB CosA SinC +SinA SinB cosC / cos A CosB SinC
=SinA ( CosB SinC + SinB cos C) + SinB CosA SinC / cos A CosB SinC
=SinA Sin[B+C] + SinB CosA SinC / cos A CosB SinC
=SinA CosA + SinB CosA SinC / cos A CosB SinC
=CosA [ SinA + SinB SinC] / cos A CosB SinC
=CosA [ SinA + SinB SinC] / cos A CosB SinC
= SinA + SinB SinC / CosB SinC
Now, multiply and divide with 2
(2sinA+2sinBsinC)/2cosBsinC
=(2sinA+cos(B-C)-cos(B+C))/(sin(B+C)-sin(B-C))
=(2sinA+cos(B-C)-cos(90-A))/(sin(90-A)-sin(B-C))
=(2sinA+cos(B-C)-sinA)/(cosA-sin(B-C))
=(sinA+cos(B-C))/(CosA-sin(B-C))
=(sin(90-(B+C))+cos(B-C))/cos(90-(B+C))-sin(B-C))
=(cos(B+c)+cos(B-C))/(sin(B+C)-sin(B-C))
=2cosBcosC/2cosBsinC
=cosC/sinC
= CotC
hope helped!
praneet u wrote hey there in middle of ur answer hehe :p
lol
lol
I see just elaborating a solution is a way to brainliest answer.
Yep ✌
Mr. yep , but I do believe in approach of the answer and not in the length of it.
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