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How do you solve #(tanx + 1)(sinx - 1) = 0#?

Answer 1

#x=-pi/4#, #x=pi/2#

#(tanx+1)(sinx-1)=0# It means either #tanx+1=0# or #sinx-1=0# Therefore we can consider both situations i.e. #tanx=-1# #x=tan^-1-1# #x=-pi/4# similarly #sinx-1=0# #sinx=1# #x=sin^-1 1# #x=pi/2#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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