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.