Math, asked by yashchampion36, 10 months ago

1/secA+tanA -1/cosA = 1/cosA - 1/secA-tanA​

Answers

Answered by ryk33
3

Answer:

LHS

1/(secA + tanA) - 1/cosA

=1/(1/cosA+sinA/cosA) -1/cosA

=cos^2A-1-sinA/cosA(1+sinA)

=1-sin^2A-1-sinA/cosA(1+sinA)

= -sinA(1+sinA)/cosA(1+sinA)

=-sinA/cosA = -tanA

RHS

1/cosA - 1/secA-tanA

=1/cosA - cosA/1-sinA

=1-sinA-cos^2A/cosA(1-sinA)

=1-sinA-1+sin^2A/cosA(1-sinA)

=-sinA(1-sinA)/cos(1-sinA)

-sinA/cosA =-tanA

Here LHS=RHS.Hence it is proved.

Answered by Anonymous
3

Answer:

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