Question

If (psin θ + qcos θ) = a and, (psin θ - qcosθ) = b, then find the value of :
$\dfrac{p + a}{q + b} + \dfrac{q - b}{p - a}$
Given Options are –
1) 1
2) 2
3) 0
4) None of these​

Answer 1

AnswEr :

$\sf{Given : pcos\theta + qsin\theta = a}\\\\\\\bigstar\:\underline{\textsf{Squaring on both sides, We get :}}\\\\\\\sf{\implies (pcos\theta + qsin\theta)^2 = a^2}\\\\\\\sf{\implies p^2cos^2\theta + q^2sin^2\theta + 2pqsin\theta cos\theta = a^2\quad\dfrac{\quad}{}[1]}$

$\rule{100}{2}$

$\sf{Given : pcos\theta - qsin\theta = b}\\\\\\\bigstar\:\underline{\textsf{Squaring on both sides, We get :}}\\\\\\\sf{\implies (pcos\theta - qsin\theta)^2 = b^2}\\\\\\\sf{\implies p^2cos^2\theta + q^2sin^2\theta - 2pqsin\theta cos\theta = b^2\quad\dfrac{\quad}{}[2]}$

$\rule{300}{1}$

$\bigstar\:\underline{\textsf{Adding both Equations [1] and [2], We get :}}$

$\sf{\implies p^2sin^2\theta + q^2cos^2\theta + \cancel{2pqsin\theta cos\theta }=a^{2} }\\\\\\\sf{ \implies p^2cos^2\theta + q^2sin^2\theta - \cancel{2pqsin\theta cos\theta} = b^2} \\ \dfrac{ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}{} \\\sf{\implies p^2sin^2\theta + q^2cos^2\theta + p^2cos^2\theta + q^2sin^2\theta = b^2 + a^2}\\\\\\\sf{\implies p^2(sin^2\theta + cos^2\theta)+q^2(sin^2\theta + cos^2\theta) = a^2 + b^2}\\\\\\\sf{\textbf{\textdagger}\:\:We\:Know : \boxed{\sf{sin^2\theta + cos^2\theta = 1}}}\\\\\\\sf{\implies p^2 + q^2 = a^2 + b^2}$

$\rule{300}{2}$

$\bigstar\:\textsf{As Per Question Now: \: \sf\dfrac{p + a}{q + b}+\dfrac{q - b}{p - a} }\\\\\\\sf{\implies \dfrac{(p + a)(p - a) + (q + b)(q - b)}{(q + b)(p - a)}}\\\\\\\sf\implies\dfrac{p^2 - a^{2} + q^2 - b^2}{(q + b)(p - a)}\\\\\\\sf\implies\dfrac{(p^2 + q^2) - (a^{2} + b^2)}{(q + b)(p - a)}\\\\\\\sf{But,\:We \:have : p^2 + q^2 = a^2 + b^2}\\\\\\\sf{\implies \dfrac{\cancel{(a^2 + b^2)}- \cancel{(a^2 + b^2)}}{(q + b)(p - a)}}\\\\\\\sf{\implies \dfrac{0}{(q + b)(p - a)}}\\\\\\\sf{\implies 0}$

⠀

$\therefore\large\boxed{\sf{Value \:of \:\boxed{\sf\dfrac{p + a}{q + b}+\dfrac{q - b}{p - a} = 0}}}$

Answer 2

Answer:

$\bold\red{Answer\:=3)0}$

$\text {Refer\:to\:attatched\:image\:for\:solution}$