Answer :

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 sin^{2}θ + cos^{2}θ = 1

⇒ cos^{2}θ = 1 – sin^{2}θ,

substituting this in the above equation, we get

Now expanding the numerator by applying the formula (a + b)^{2} = a^{2} + b^{2} + 2ab, we get

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

⇒ 1 + x^{2} + 2x = p^{2}(1 – x^{2})

⇒ 1 + x^{2} + 2x = p^{2} – p^{2}x^{2}

⇒ 1 + x^{2} + 2x – p^{2} + p^{2}x^{2} = 0

⇒ (1 + p^{2})x^{2} + 2x + (1 – p^{2}) = 0

Comparing this with standard equation, i.e., Ax^{2} + Bx + C = 0, we get

A = (1 + p^{2}), B = 2, C = (1 – p^{2})

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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