(tan＾2α－cot＾2α)/(sin＾2α－cos＾2α)＝sec＾2α•csc＾2α
问题描述：
(tan＾2α－cot＾2α)/(sin＾2α－cos＾2α)＝sec＾2α•csc＾2α
答
？，恒等式；
等号左端=[(sin²α/cos²α)-(cos²α/sin²α)] /(sin²α-cos²α)
=[(sin²αsin²α-cos²αcos²α]/[(sin²α-cos²α)(sin²α•cos²α)]
=[(sin²α-cos²α)(sin²α+cos²)]/[(sin²α-cos²α)(sin²αcos²α)]
=1/(sin²αcos²α)=sec²α•csc²α=等式右端；
答
(tan²α-cot²α)/(sin²α-cos²α)=(sin²α/cos²α-cos²α/sin²α)/(sin²α-cos²α)=(sin⁴α-cos⁴α)/[sin²αcos²α(sin²α-cos²...