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Q-1 समीर: कथं चलति?
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Answer 1

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

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

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

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

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

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