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Could U pleaseeeeeeeeee prove this?

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Sin^2x.cos^2x.tan^2x+(cos^2x.cot^2x)/(1+cot^2x)=1-2sin^2x.cos^2x
asked Mar 26, 2013 in TRIGONOMETRY by M Gravity Rookie

1 Answer

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Sin^2x.cos^2x.tan^2x+(cos^2x.cot^2x)/(1+cot^2x)

Recall the quotient trignometric identity

cotx=cosx/sinx

=Sin^2x.cos^2x.tan^2x+(cos^2x.cos^2x/sin^2x)/(1+(cos^2x/sin^2x))

=Sin^2x.cos^2x.tan^2x+(cos^2x.cos^2x/sin^2x)/((sin^2x+cos^2x)/sin^2x)

Recall pythagoras theorem of trignometry

sin^2x+cos^2x=1

=Sin^2x.cos^2x.tan^2x+(cos^2x.cos^2x/sin^2x)/(1/sin^2x)

Simplify

=Sin^2x.cos^2x.tan^2x+(cos^2x.cos^2x)

Recall the quotient trignometric identity

tanx=sinx/cosx

=Sin^2x.cos^2x.sin^2x/cos^2x.+(cos^2x.cos^2x)

=Sin^2xsin^2x+(cos^2x.cos^2x)

=(sin^2x)^2+(cos^2x)^2

Recall the formulae a^2+b^2=(a+b)^2-2ab

=(sin^2x+cos^2x)^2-2sin^2xcos^2x

Recall pythagoras theorem of trignometry

sin^2x+cos^2x=1

=1-2sin^2xcos^2x

 

answered Mar 27, 2013 by bradely Mentor
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