Question

prove that 1/sec-tan -1/cos = 1/cos -1/sec+tan

Answer 1

on solving both sides separately, tan theta will come on both sides..
hence LHS=RHS.. ( PROVED)
hope it helps..
pls mark brainliest...

Answer 2

Given:t 1/sec-tan -1/cos = 1/cos -1/sec+tan

To Proof:1/sec-tan -1/cos = 1/cos -1/sec+tan

Proof:

1/sec - tan = 1/(1/cos) - tan = cos - tan

Next, we can use the identity tan^2 + 1 = sec^2 to simplify the expression:

cos - tan = cos - tan / (1 + tan^2) * tan = cos - tan^2 / (1 + tan^2)

Finally, we can use the identity 1 + tan^2 = sec^2 to simplify further:

cos - tan^2 / (1 + tan^2) = cos - tan^2 / sec^2 = cos / sec^2 - tan^2 / sec^2 = 1 / sec - tan^2 / sec

This is equivalent to the right-hand side of the equation:

1 / sec - tan^2 / sec = 1 / sec - tan / cos = 1 / cos - 1 / sec + tan

Therefore, we can conclude that:

1/sec - tan = 1/cos - 1/sec + tan.

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