Question

If p sin theta +q cos theta =a and p cos theta -q cos theta =b then p+a/q+b+q-b/p-a is equal to?

Answer 1

Answer:

$\frac{p+a}{q+b}+\frac{q-b}{p-a}=0$

Step-by-step explanation:

Formula used:

$cos^2\theta+sin^2\theta=1$

$(a+b)^2=a^2+b^2+2ab$

$(a-b)^2=a^2+b^2-2ab$

Given:

$p\:sin\theta+q\:cos\theta=a$......(1)

$p\:cos\theta-q\:sin\theta=b$........(2)

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

$(p\:sin\theta+q\:cos\theta)^2+(p\:cos\theta-q\:sin\theta)=a^2+b^2$

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

$p^2\:sin^2\theta+q^2\:cos^2\theta+p^2\:cos^2\theta+q^2\:sin^2\theta=a^2+b^2$

$p^2(sin^2\theta+cos^2\theta)+q^2(cos^2\theta+sin^2\theta)=a^2+b^2$

$p^2(1)+q^2(1)=a^2+b^2$

$p^2+q^2=a^2+b^2$............(3)

Now,

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

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

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

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

=$\frac{(a^2+b^2)-(a^2+b^2)}{(q+b)(p-a)}$  ( using (3) )

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

=$0$

Answer 2

Answer:

H

Step-by-step explanation: