Question #223094

Prove the following trig identities

sin 2x + cos 2x / sin x + cos x = 2cosx

2sinxcosx / cos4x-sin4x = tan2x


1
Expert's answer
2021-09-10T00:14:19-0400

(b)


2sinxcosxcos4xsin4x=tan2x\frac{2sinxcosx}{cos^4x-sin^4x}=tan2x

taking LHS

=sin2x(cos2x+sin2x)(cos2xsin2x)=sin2xcos2xsin2x=sin2xcos2x=tan2x=\frac{sin2x}{(cos^2x+sin^2x)(cos^2x-sin^2x)}\\=\frac{sin2x}{cos^2x-sin^2x}\\=\frac{sin2x}{cos2x}\\=tan2x

(a)

sin2x+cos2xcosx+sinx=2cosx\frac{sin2x+cos2x}{cosx+sinx}=2cosx

taking LHS

=2sinxcosx+2cos2x1cosx+sinx=\frac{2sinxcosx+2cos^2x-1}{cosx+sinx}

2cosx1sinx+cosx2cosx- \frac{1}{sinx+cosx}

so sinx +cosx cannnot be zero so this is not solved further


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