Question #350e8

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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