Question

Soni was trying hard to prove $asin\theta +bcos\theta =\pm \sqrt{a^2+b^2-c^2}$. When $acos\theta -bsin\theta =c$. Her classmate Swati gave her a hint of squaring both sides of $acos\theta -bsin\theta =c$ and proceed further.
Following her hint, Soni was able to solve the problem and she thanks Swati for this hint. Write the solution of the problem.

Answer 1

We have, $(acos\theta -bsin\theta )=c$

$=(acos\theta -bsin\theta {)}^2=c^2$ (squaring both sides)

$=a^{2}co{s}^2\theta +b^{2}si{n}^2\theta -2absin\theta \times cos\theta =c^2$

$=a^2(1-si{n}^2\theta )+b^2(1-co{s}^2\theta )-2absin\theta \times cos\theta =c^2$

$=$ $a^{2}si{n}^2\theta +b^{2}co{s}^2\theta +2absin\theta \times cos\theta$=$a^2+b^2-c^2$

$=(asin\theta +bcos\theta {)}^2=a^2+b^2-c^2$

$=(asin\theta +bcos\theta )=\pm \sqrt{a^2+b^2-c^2}$