Question #350e8

Question

1110 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-13T22:59:29

see below

#sin(pi-x)-tan(pi+x)=sin x (cosx-1)/cosx#

Use the formulas: #sin(A-B)=sin A cos B-cos A sin B#

#sin(A+B)=sin A cos B+cos A sin B#

#cos(A-B)=cos A cos B+ sin A sin B#

Left Hand Side: #=sin(pi-x)-tan(pi+x)#

#=sin(pi-x)-(sin(pi+x))/cos(pi+x)#

#=(sin pi cos x - cos pi sin x) -((sin pi cos x + cos pi sin x)/(cos pi cos x - sin pi sin x))#

#=(0* cos x - (-1)* sin x) -((0*cos x + (-1) sin x)/(-1*cos x + 0* sin x))#

#=sin x-((cancel-sinx)/(cancel-cos x))#

#=sinx-sinx/cosx#

#=(sinxcosx-sinx)/cos x#

#=sinx (cosx-1)/cosx#

#:. =# Right Handside

Answer 2 · 2016-12-14T01:46:22

#LHS=sin(pi-x)-tan(pi+x)#

#=sinx-tanx#

#=sinx-sinx/cosx#

#=sinx(1-1/cosx)#

#=sinx((cosx-1)/cosx)=RHS#

Proved