已知sin²γ=sin²α-sinαcosαtan(α-β),求证tan²γ=tanαtanβ急...
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已知sin²γ=sin²α-sinαcosαtan(α-β),求证tan²γ=tanαtanβ急...
已知sin²γ=sin²α-sinαcosαtan(α-β),求证tan²γ=tanαtanβ
急...
已知sin²γ=sin²α-sinαcosαtan(α-β),求证tan²γ=tanαtanβ急...
∵sin²γ=sin²α-sinαcosαtan(α-β)①
∴cos²γ=1-sin²γ=1-[sin²α-sinαcosαtan(α-β)】
=1-sin²α+sinαcosαtan(α-β)
cos²γ =cos²α+sinαcosαtan(α-β) ②
①/②:
tan²γ
=sin²γ/cos²γ
=[sin²α-sinαcosαtan(α-β)]/[cos²α+sinαcosαtan(α-β)]
=[sin²α/cos²α-sinα/cosα*tan(α-β)]/[1+tanαtan(α-β)]
=[tan²α-tanα*tan(α-β)]/[1+tanαtan(α-β)]
=tanα[tanα-tan(α-β)]/[1+tanαtan(α-β)]
=tanαtan[α-(α-β]
=tanαtanβ