Question

prove sinA + sinB sinc/cos B sin C= cos c / sin c

Answer 1

GIVEN A+B+C=90

TO PROVE  SINA + SINBSINC / COSBCOSC = COSC/SINC

B+C = 90-A

COS(B+C) = COS(90-A)

COSBCOSC - SINBSINC = SINA

COSBCOSC = SINA+SINBSINC

COSC = SINA+ SINBSINC / COSB

DIVIDING BOTH SIDES BY SINC

COSC / SINC = SINA + SINBSINC / COSB SINC

HENCE PROVED.

Answer 2

you didn't mention about relation between A, B and C .
I think , Question is ,

A + B + C = 90°
then,
prove that :
(sinA + sinB.sinC)/cosB.sinC = cosC/sinC

solution :-

A + B + C = 90°
B + C = 90° - A
take both side cos

cos( B + C) = cos(90° - A)

[ use, formula, cos(x + y)= cosx.cosy-sinx.siny and cos(90°-x) = sinx ]

cosB.cosC - sinB.sinC = sinA

cosB.cosC = sinA + sinB.sinC

divide both sides , with cosB.sinC

cosB.cosC/cosB.sinC = (sinA+sinB.sinC)/cosB.sinC

cosB/sinC = (sinA + sinB.sinC)/cosB.sinC

hence, proved ////