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Answer to Question #10053 in Trigonometry for archana meena

Question #10053
1/secA+tanA=1-secA/cosA







Expert's answer
This is not identity.
If we have 1/(secA+tanA)=(1-secA)/cosA then it is equal
to cosA/(sinA+cosA)=(cosA-1)/cos^2(A)
If we put A=60 degrees then cosA=1/2
and sinA=sqrt(3)/2.
In left part of identity we'll get 1/(sqrt(3)+1) and for
the right part -2. They are not equal.

If we omit parenthesis
1/secA+tanA=1-secA/cosA then it's equivalent to (cos^2(A)+sinA)/cosA=
-sin^2(A)/cos^2(A).
Similar way we'll get again not equal parts.

So,
it is error in given statement.

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