Question

求证sin*2 αtanα+cos*2 αcotα+2sinαcosα-cotα=tanα

Answer 1

sin²αtanα+cos²αcotα+2sinαcosα-cotα=tanα
左边
=sin³α/cosa +cos³α/sina+2sinαcosα-cosα/sina
=(sin^4a+cos^4α)/﹙sinacosa﹚+﹙2sin²αcosα-cosα﹚/sina
=(sin^4a+cos^4α)/﹙sinacosa﹚+﹙好蠢2sin²α-1)cosα/sina
=(sin^4a+cos^4α)/﹙sinacosa﹚-cos2αcos²α/﹙sinacosa﹚
=(sin^4a+cos^4α-2cos^4a+cos²α﹚/﹙sinacosa﹚
=(sin^4a-cos^4α+cos²α)/﹙sinacosa﹚
=[(sin²a-cos²α)(sin²孝御a﹢cos²α)+cos²α]/﹙sinacosa﹚
=sin²a/﹙sinacosa﹚
=sina/巧袜岩cosa
=tanα
右边
=tanα
∴sin² αtanα+cos² αcotα+2sinαcosα-cotα=tanα

Answer 2

＝sinα/禅拆cosα＋cosα/sinα ＝tanα＋cotα ＝右边。 sin^2α×tanα＋衫肆cos^2α×cotα＋或袭轿2sinα×cosα =2sinacosaxsina/cosa+(cosacosa-sina