If A,B,C are angles of a triangle and none of them is equal to
then prove that
tanA + tanB + tanc = tanA. tanB. tanC.
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Answer:
ANSWER
→ (i) tanA+tanB++tanC
=
cosAcosB
sinAcosB+sinB+cosA
+tanC
cosBcosB
sin(A+B)
+
cosC
sinC
=
cosBcosA
sinC
+
cosC
sinC
(sinA−C=sinC)
=sinC(
cosAcosBcosC
cosC+cosAcosB
)
=sinC(
cosAcosBcosC
cos(π−(A+B))+cosAcosB
)
=sinC(
cosAcosBcosC
cosAcosB−cos(A+B)
)
=sinC(
cosAcosBcosC
cosAcosB−cosAcosB+sinAsinB
)
=
cosAcosBcosC
sinAsinBsinC
=tanAtanBtanC
(ii) cotAcotB=
tanAtanB
1
=
tanA+tanB+tanC
tanC
cotBcosC=
∑tanA
tanA
,cosAcosC=
∑tanA
tanB
⇒ Sum=
tanA+tanB+tanC
tanA+tanB+tanC
=1
Answered By
Pragyan
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