Question

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

a+b+c=90°

Answer 1

Hi bro!!!!

( sinA + sinB sinc)/cos B sin C

To solve this question, you should know transformation formulas..I will mention those formulas where I used in square brackets..

Firstly multiply and divide with 2

(2sinA+2sinBsinC)/2cosBsinC

★[2sinBsinC=cos(B-C)-cos(B+C)]★

★[2cosBsinC=sin(B+C)-sin(B-C)]★

→(2sinA+cos(B-C)-cos(B+C))/(sin(B+C)-sin(B-C))
since B+C =90-A

we get,

=(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))

★[cos(B+c)+cos(B-C)=2cosBcosC]★

★[sin(B+C)-sin(B-C)=2cosBsinC]★

=2cosBcosC/2cosBsinC

=cosC/sinC

hence proved..

I hope this will help u ;)

Answer 2

Hi ,

here we use trigonometric transformations .


LHS = ( sinA + sinBsinC )/ cosBsinC

multiply numerator and Denominator with 2

= ( 2sinA + 2sinBsinC)/2cosBsinC

{Given A+B+C = 90

A = 90 - ( B+ C)

sinA = sin[ 90 - ( B+ C)]

sinA = cos ( B+C) }

= [2cos(B+C)+ cos( B-C) - cos(B+C)]/2cosBsinC

{ since 2sinBsinC = cos(B-C) - cos( B+C) }

= [ cos(B+C) +cos(B-C)]/2cosBsinC

= ( 2cosBcosC)/ (2cosBsinC)

{ since cos( B+C) + cos(B-C) = 2cosBcosC }

after cancellation

=cosC/ sinC

= RHS

I hope this helps you.

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