# 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.

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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