prove that secA(1-sinA)-2tanA=1

secA-tanA-2tanA=1

secA-3tanA=1

To prove secA(1-sinA)-2tanA = 1, we need to simplify the left side of the equation and show that it is indeed equal to 1.

Let's start by analyzing each term individually:

1. secA:
The reciprocal of cosine is secant, given by secA = 1/cosA.

2. 1 - sinA:
This is a simple subtraction of sinA from 1.

3. -2tanA:
The negative two (-2) is multiplied by tanA, where tanA = sinA/cosA.

Now, let's substitute these values back into the equation and simplify it:

secA(1 - sinA) - 2tanA
= (1/cosA)(1 - sinA) - 2(sinA/cosA)
= (1 - sinA)/cosA - 2sinA/cosA

To combine the terms, we need a common denominator, which is cosA. We can achieve this by multiplying the first term by cosA/cosA and the second term by (2cosA)/cosA:

= [(1 - sinA)(cosA) - 2sinA(2cosA)] / cosA
= [cosA - sinA(cosA) - 4sinA(cosA)] / cosA
= [cosA - cosAsinA - 4cosAsinA] / cosA
= [cosA - 5cosAsinA] / cosA

Now, we can factor out cosA from the numerator:

= cosA(1 - 5sinA) / cosA
= 1 - 5sinA

And we have arrived at 1 - 5sinA, which is not equal to 1. Therefore, the equation secA(1-sinA)-2tanA = 1 is incorrect.