**Given sin θ = p/q, find cos θ + sin θ in terms of p and q.**

**Answer :**

Given that sin θ = p/q

AB/AC = p/q

Let AB = px

And then AC = qx

In right angled triangle ABC

AC^{2} = AB^{2} + BC^{2}

⇒ BC^{2} = AC^{2} – AB^{2}

⇒ BC^{2} = q^{2}x^{2} – p^{2}x^{2}

⇒ BC^{2} = (q^{2} – p^{2})x^{2}

⇒ BC = √( q^{2} – p^{2})x

In right angled triangle ABC,

cos θ = base/hypotenuse

= BC/AC

= √( q^{2} – p^{2})x/qx

= √( q^{2} – p^{2})/q

Now,

Sin θ + cos θ = p/q + √(q^{2}–p^{2})/q

= [p + √(q^{2}–p^{2})]/q

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