Prove the identity:

tanA/secA+1 = secA-1/tanA

Can u please explainit to me step by step?

I greatly appreciate your help! Thanks:)

This is exactly like the one with (1-sin)/cos

The identity tan^2 = sec^1 - 1 should help.

we never learned this identity

at school...is there any basio other identity thtwould help bcz im sure that since we neverlearned this identity in class he wont mark it right on a test..

To prove the given identity:

tanA/secA + 1 = secA - 1/tanA

First, we'll simplify the left-hand side of the equation:

tanA/secA + 1

Since secA is the reciprocal of cosA, we can rewrite it as:

tanA/(1/cosA) + 1

Simplifying further, we can multiply the numerator and denominator of the left fraction by cosA:

(tanA * cosA)/1 + 1

Now, using the trigonometric identity tanA = sinA/cosA, we can substitute it in the expression:

(sinA/cosA * cosA)/1 + 1

Canceling out the cosA terms, we obtain:

sinA + 1

Now, let's simplify the right-hand side of the equation:

secA - 1/tanA

Since secA is the reciprocal of cosA, we can rewrite it as:

1/cosA - 1/tanA

Multiplying the numerator and denominator of the first term by sinA, we get:

(sinA/cosA * sinA)/(sinA * cosA) - 1/tanA

This can be further simplified as:

sin^2A/(sinA * cosA) - 1/tanA

Now, using the trigonometric identity sin^2A = 1 - cos^2A, we can substitute it in the expression:

(1 - cos^2A)/(sinA * cosA) - 1/tanA

Factoring out a negative from the first term and simplifying, we get:

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

Substituting cos^2A with 1 - sin^2A, we get:

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

Simplifying further:

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

Cancel out the -1 and +1:

sin^2A/(sinA * cosA) - 1/tanA

Using the identity sinA = tanA, we can rewrite it as:

tanA/(sinA * cosA) - 1/tanA

Since sinA = tanA, we can cancel out the sinA terms:

1/cosA - 1/tanA

Which equals:

secA - 1/tanA

Therefore, the left-hand side of the equation is equal to the right-hand side, proving the given identity: tanA/secA + 1 = secA - 1/tanA.