Question

PLs help me with this question

Answer 1

$\sf{Given : psin\theta + qcos\theta = a}\\\\\\\textsf{Squaring on both sides, We get :}\\\\\\\sf{\implies (psin\theta + qcos\theta)^2 = a^2}\\\\\\\sf{\implies p^2sin^2\theta + q^2cos^2\theta + 2pqsin\theta cos\theta = a^2\;------\;[1]}$

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

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

$\sf{\implies p^2sin^2\theta + q^2cos^2\theta + 2pqsin\theta cos\theta + p^2cos^2\theta + q^2sin^2\theta - 2pqsin\theta cos\theta = b^2 + a^2}$

$\sf{\implies p^2sin^2\theta + q^2cos^2\theta + p^2cos^2\theta + q^2sin^2\theta = b^2 + a^2}\\\\\\\sf{\implies p^2(sin^2\theta + cos^2\theta) + q^2(sin^2\theta + cos^2\theta) = a^2 + b^2}\\\\\\\sf{\bigstar\;\;We\;know\;that : \boxed{\sf{sin^2\theta + cos^2\theta = 1}}}\\\\\\\sf{\implies p^2 + q^2 = a^2 + b^2}$

$\sf{Now, Consider :\;\dfrac{p + a}{q + b} + \dfrac{q - b}{p - a}}\\\\\\\textsf{Taking L.C.M of above Fractions, We get :}\\\\\\\sf{\implies \dfrac{(p + a)(p - a) + (q + b)(q - b)}{(q + b)(p - a)}}\\\\\\\implies \dfrac{p^2 - a^2 + q^2 - b^2}{(q + b)(p - a)}\\\\\\\sf{But,\;We\;found\;that : p^2 + q^2 = a^2 + b^2}\\\\\\\sf{\implies \dfrac{a^2 + b^2 - a^2 - b^2}{(q + b)(p - a)}}\\\\\\\sf{\implies \dfrac{0}{(q + b)(p - a)}}\\\\\\\sf{\implies 0}$

Answer 2

$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,

$psin\: \theta + qcos\: \theta=a$ ———[1]

$pcos\: \theta - qsin\: \theta=b$ ———[2]

$\dfrac{p+a}{q+b}+\dfrac{q-b}{p-a} = ?$ ———[3]



$\underline{\huge{\textsf{Step-1:}}}$
Simplify LHS of [3]

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

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



$\underline{\huge{\textsf{Step-2:}}}$
Using the Identity $\mathbf{(a+b)(a-b)= a^{2}-b^{2}}$

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

On Substituting,

$\implies \dfrac{(p^{2}-a^{2})+(q^{2}-b^{2})}{(q+b)(p-a)}$

$\implies \dfrac{(p^{2}-a^{2})+(q^{2}-b^{2})}{(q+b)(p-a)}$

$\implies \dfrac{(p^{2}+q^{2})-(a^{2}+b^{2})}{(q+b)(p-a)}$ ———[4]



$\underline{\huge{\textsf{Step-3:}}}$
Square on Both sides of eq. [1] & [2]

$\implies (psin\: \theta + qcos\: \theta)^{2}=a^{2}$

$\implies (pcos\: \theta - qsin\: \theta)^{2}=b^{2}$



$\underline{\huge{\textsf{Step-4:}}}$
Using the Identities
$\mathbf{(a+b)^{2} = a^{2}+b^{2}+2ab}$
$\mathbf{(a-b)^{2}= a^{2}+b^{2}-2ab}$

$\implies p^{2}sin^{2}\: \theta+q^{2}sin^{2}\: \theta+2pqsin\: \theta\, cos\: \theta = a^{2}$ ———[5]

$\implies p^{2}cos^{2}\: \theta+q^{2}sin^{2}\: \theta - 2pqsin\: \theta\, cos\: \theta = b^{2}$ ———[6]



$\underline{\huge{\textsf{Step-5:}}}$
Add Equations [5] & [6] to get $a^{2}+b^{2}$

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

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



$\underline{\huge{\textsf{Step-6:}}}$
Using the Identity $\mathbf{sin^{2}\: \theta+cos^{2}\: \theta= 1}$

On Substituting

$\implies a^{2}+b^{2} = p^{2}(1)+q^{2}(1)$

$\implies a^{2}+b^{2} = p^{2}+q^{2}$ ———[7]



$\underline{\huge{\textsf{Step-7:}}}$
Substitute [7] in [4]

$\implies \dfrac{(p^{2}+q^{2})-(p^{2}+q^{2})}{(q+b)(p-a)}$

$\implies \dfrac{0}{(q+b)(p-a)}$

$\implies 0$

$\therefore$

$\blacksquare \: \mathbf{\dfrac{p+a}{q+b}+\dfrac{q-b}{p-a} = \underline{\large{0}}}$