How do you simplify the expression #tan^2t(csc^2t-1)#?

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Answer 1 · 2017-01-20T13:37:54

#1.#

We use the Identity #csc^2t=1=cot^2t# to get,

#tan^2t(csc^2t-1)=(tan^2t)(cot^2t)=1#.

Otherwise , substituting #sint/cost" for "tant" and "1/sint" for "csct#,

The L.H.S.#=(sin^2t/cos^2t){1/sin^2t-1}#

#={cancel(sin^2t)(1-sin^2t)}/{(cos^2t)cancel(sin^2t)}#

#=cos^2t/cos^2t#

#=1.#