Question

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Correct detailed answer is needed.
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Answer 1

Answer:

Refer Image

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

We have,

$\bf{sin(\theta)=\dfrac{p}{q}}$

Now,

$\sf{cos(\theta)=\sqrt{1-sin^2(\theta)}=\sqrt{1-\left(\dfrac{p}{q}\right)^2}}$

$\sf{\implies\,cos(\theta)=\sqrt{1-\dfrac{p^2}{q^2}}}$

$\sf{\implies\,cos(\theta)=\sqrt{\dfrac{q^2-p^2}{q^2}}}$

$\sf{\implies\,cos(\theta)=\dfrac{\sqrt{q^2-p^2}}{q}}$

So,

$\sf{\blue{sin(\theta)+cos(\theta)}}$

$\sf{=\dfrac{p}{q}+\dfrac{\sqrt{q^2-p^2}}{q}}$

$\sf{=\dfrac{p+\sqrt{q^2-p^2}}{q}}$