If sec θ + tan θ

Given, sec θ + tan θ = p

But we know ,

substituting these values in the given equation, we get

Now we will square on both sides, we get

Now we know sin2θ + cos2θ = 1
cos2θ = 1 – sin2θ,
substituting this in the above equation, we get

Now expanding the numerator by applying the formula (a + b)2 = a2 + b2 + 2ab, we get

Now let x = sin θ, so the above equation becomes

1 + x2 + 2x = p2(1 – x2)

1 + x2 + 2x = p2 – p2x2

1 + x2 + 2x – p2 + p2x2 = 0

(1 + p2)x2 + 2x + (1 – p2) = 0

Comparing this with standard equation, i.e., Ax2 + Bx + C = 0, we get

A = (1 + p2), B = 2, C = (1 – p2)

So the value of x of a quadratic equation can be found by using the formula,

Now substituting the corresponding values, we get

So the two possibilities are,

But we had assumed x = sin θ, substituting back we get

And we know,

Substituting the value of sin θ, we get,

These are the required value of cosec θ.

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