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\sin^2(42\degree +\alpha ) +\cot(25\degree +\beta )\cdotp \cot (\beta -65\degree )+\sin^2(48\degree -\alpha ) =

  1. 0

Collected in the board: Trigonometry

Steven Zheng Steven Zheng posted 2 months ago


Answer

  1. Apply Pythagorean Identity and cofucntion identity


    \sin^2(42\degree +\alpha ) +\cot(25\degree +\beta )\cdotp \cot (\beta -65\degree )+\sin^2(48\degree -\alpha )

    =\sin^2(42\degree +\alpha ) +\sin^2[90\degree -(\alpha+42\degree ) ]+\cot(90\degree-65\degree +\beta )\cdotp \cot (\beta -65\degree )

    =\sin^2(42\degree +\alpha ) +\cos^2(42\degree +\alpha ) -\tan(\beta-65\degree )\cdotp \cot (\beta -65\degree )

    =1-1=0

Steven Zheng Steven Zheng posted 2 months ago


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