Filling Question

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

Collected in the board: Trigonometry

Steven Zheng Steven Zheng posted 1 year 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 1 year ago

Scroll to Top