Question

Iif acos theta - bsin theta =c
Then prove that asin theta +bcos theta= +_ root a^2 +b^2+c^2

Answer 1

I hope you will understand step by step explanation

Answer 2

Answer:

Proved.

Step-by-step explanation:

                                  $\sf aCos \ \theta - bSin \ \theta = c\\\\\text{\bf Square both sides},\\\\ (aCos \ \theta - bSin \ \theta)^2 = c^2$

Use Identity (a - b)² = a² + b² - 2ab

       $\sf a^2Cos^2 \ \theta + b^2Sin^2 \ \theta -2bSin \ \theta*aCos \ \theta= c^2\\\\ a^2Cos^2 \ \theta + b^2Sin^2 \ \theta - 2abSin \ \theta*Cos \ \theta = c^2$

We know that,

    $\sf \boxed{\bf Cos^2 \ \theta = 1 -Sin^2 \ \theta}\\\\ \boxed{\bf Sin^2 \ \theta = 1 - Cos^2 \theta}$

    $\sf a^2(1 - Sin^2 \ \theta) + b^2(1 - Cos^2 \ \theta)-2abSin \ \theta*Cos \ \theta = c^2\\\\a^2 - a^2Sin^2 \ \theta + b^2- b^2Cos^2 \ \theta - 2abSin \ \theta*Cos \ \theta = c^2$

                $\sf -a^2Sin^2 \ \theta - b^2 Cos^2 \ \theta - 2ab SIn \ \theta*Cos \ \theta =c^2 - a^2 - b^2\\\\\text{\bf Mulitply the entire equation by (-1)}\\\\ a^2Sin^2 \ \theta + b^2Cos^2 \ \theta + 2abSin \ \theta *Cos \ \theta = a^2 + b^2 - c^{2}$

                                  $\sf (aSin \ \theta +bCos \ \theta)^2 = a^2 + b^2 - c^2\\\\\text{Take square root,}$

                                   $\sf aSin \ \theta + bCos \ \theta = \± \sqrt{a^2 + b^2 - c^2}$

Hence proved.