Question

solve it for 50 points spam will be reported​

Answer 1

$\huge\blue\star\underbrace\mathtt{\underline{\underline\red{Answer}}}$

Formula used:

$\sf \gray{cos^2\theta+sin^2\theta=1cos}$

$\sf \pink{(a+b)^2=a^2+b^2+2ab(a+b)}$

$\sf \blue{(a-b)^2=a^2+b^2-2ab(a−b)}(a−b)$

Given:

$\sf \orange{p\:sin\theta+q\:cos\theta=a}$

(1)

$\sf \red{p\:cos\theta-q\:sin\theta=b}$

(2)

squaring and adding (1) and (2), we get

$\sf \pink{(p\:sin\theta+q\:cos\theta)^2+(p\:cos\theta-q\:sin\theta)=a^2+b^2(psinθ+qcosθ)}(psinθ+qcosθ)$

$\sf \gray{p^2\:sin^2\theta+q^2\:cos^2\theta+2pq\:cos\theta\:sin\theta+p^2\:cos^2\theta+q^2\:sin^2\theta-2pq\:cos\theta\:sin\theta=a^2+b^2p}$

$\sf \blue{p^2\:sin^2\theta+q^2\:cos^2\theta+p^2\:cos^2\theta+q^2\:sin^2\theta=a^2+b^2p}$

$\sf \purple{p^2(sin^2\theta+cos^2\theta)+q^2(cos^2\theta+sin^2\theta)=a^2+b^2p}$

$\bf \: p^2(1)+q^2(1)=a^2+b^2pp$

$\sf \:p^2+q^2=a^2+b^2pp$

.(3)

Now,

$\bf \blue{\frac{p+a}{q+b}+\frac{q-b}{p-a}}$

$\cal \purple{=\frac{(p+a)(p-a)+(q+b)(q-b)}{(q+b)(p-a)}}$

$\rm \orange{=\frac{p^2-a^2+q^2-b^2}{(q+b)(p-a)}}$

$\tt \green{=\frac{(p^2+q^2)-(a^2+b^2)}{(q+b)(p-a)}}$

$\sf \red{=\frac{(a^2+b^2)-(a^2+b^2)}{(q+b)(p-a)}}$

( using (3))

$\sf\pink{\boxed{=\frac{0}{(q+b)(p-a)}}}$

Answer 2

Refer from the attachment provided...

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