Verify. (SinBCosB / TanB) = 1-sin^2 B?

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2007-10-24 18:35:15 · 4 answers · asked by senorita_otono 2 in Science & Mathematics ➔ Mathematics

sin B/ tan B = cos B

so LHS = cos^2 B = 1 - sin ^2 B
proved

2007-10-24 18:38:53 · answer #1 · answered by Mein Hoon Na 7 · 0⤊ 0⤋

SinBCosB/TanB = sinBCosB/{SinB/CosB}
= Cos^2B = 1 - Sin^2B

2007-10-24 18:46:16 · answer #2 · answered by Snoopy 3 · 0⤊ 0⤋

(sinBcosB/TanB)=1-sin^2B
take note that tanB=sinB/cosB
(sinBcosB)/sinBcosB=1-sin^2B
cos^2B=1-sin^2B
1-sin^2B=1-sin^2B

2007-10-24 18:46:42 · answer #3 · answered by _sTepH_ 3 · 0⤊ 0⤋

sin B cos B
--------------
sin B
------
cos B

sin B cos Â² B
----------------
sin B

cos Â² B = 1 - sin Â² B

2007-10-24 19:41:52 · answer #4 · answered by Como 7 · 0⤊ 0⤋