Question

sin(50+ø) -- cos (40-- ø)+ tan 1 tan 10 tan 20 tan 70 tan 80 tan 89=1

plzz prove this
I will mark as brainlist......its urgent..plzz​

Answer 1

Explanation:

sin(50+∅)-cos(40-∅)+tan1.tan10.tan20.tan70.tan80.tan89

=cos40+∅ - cos40+∅ + cot89.tan89.cot80.tan80.tan70.cot70

=1. [all the above get eliminated]

Answer 2

Hey Mate

Here's Your Answer!!!!

sin(50+A)-Cos(40-A)+tan1 tan10 tan 20 tan70tan 80 tan 89=1

=sin(50+A)-cos[90-50-A] + tan(90-89)tan(90-80)tan(90-70)tan70tan80tan89

=sin(50+A)-sin(50+A). +cot89cot80cot70tan70tan80tan89

=(cot 89 * tan89 )(cot80*tan80)(cot 70 * tan 70)

From the law that

TanA=1/cotA

The above left over will get cancelled and will be equal to 1

Hence LHS=RHS

thank you

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