QUESTION

# If ${\text{acos}}\theta {\text{ - bsin}}\theta {\text{ = c}}$ , then ${\text{asin}}\theta {\text{ + bcos}}\theta$ is equal to ________________. A. $\pm \sqrt {{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}}$B. $\pm \sqrt {{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - }}{{\text{c}}^2}}$C. $\pm \sqrt {{{\text{c}}^2}{\text{ - }}{{\text{a}}^2}{\text{ - }}{{\text{b}}^2}}$D. None of these

Hint: To solve this question we will use the property ${\text{si}}{{\text{n}}^2}{\text{x + co}}{{\text{s}}^2}{\text{x = 1}}$ in the given condition to solve the problem.

Now, we will use the property ${\text{si}}{{\text{n}}^2}{\text{x + co}}{{\text{s}}^2}{\text{x = 1}}$ in the given condition ${\text{acos}}\theta {\text{ - bsin}}\theta {\text{ = c}}$ and then we will find the value of ${\text{asin}}\theta {\text{ + bcos}}\theta$. Such questions can be solved only if we do some mathematical operations on the given condition. Here in this question we can clearly see from the option that the we have to perform the square operation on the given condition ${\text{acos}}\theta {\text{ - bsin}}\theta {\text{ = c}}$ to get the correct answer.
Squaring both sides of the equation ${\text{acos}}\theta {\text{ - bsin}}\theta {\text{ = c}}$, we get
${{\text{(acos}}\theta {\text{ - bsin}}\theta )^2}{\text{ = }}{{\text{c}}^2}$
$\Rightarrow$ ${{\text{a}}^2}{\text{co}}{{\text{s}}^2}\theta {\text{ + }}{{\text{b}}^2}{\text{si}}{{\text{n}}^2}\theta {\text{ - 2absin}}\theta {\text{cos}}\theta {\text{ = }}{{\text{c}}^2}$
$\Rightarrow$ ${\text{2absin}}\theta {\text{cos}}\theta {\text{ = }}{{\text{a}}^2}{\text{co}}{{\text{s}}^2}\theta {\text{ + }}{{\text{b}}^2}{\text{si}}{{\text{n}}^2}\theta {\text{ - }}{{\text{c}}^2}$ ……. (1)
Now, squaring ${\text{asin}}\theta {\text{ + bcos}}\theta$, we get
${{\text{(asin}}\theta {\text{ + bcos}}\theta )^2}{\text{ = }}{{\text{a}}^2}{\text{si}}{{\text{n}}^2}\theta {\text{ + }}{{\text{b}}^2}{\text{co}}{{\text{s}}^2}\theta {\text{ + 2absin}}\theta {\text{cos}}\theta$
Now, putting value of $2{\text{absin}}\theta {\text{cos}}\theta$ from equation (1) in the above equation, we get
${{\text{(asin}}\theta {\text{ + bcos}}\theta )^2}{\text{ = }}{{\text{a}}^2}{\text{si}}{{\text{n}}^2}\theta {\text{ + }}{{\text{b}}^2}{\text{co}}{{\text{s}}^2}\theta {\text{ + }}{{\text{a}}^2}{\text{co}}{{\text{s}}^2}\theta {\text{ + }}{{\text{b}}^2}{\text{si}}{{\text{n}}^2}\theta {\text{ - }}{{\text{c}}^2}$
Taking ${{a}^2}$ and ${{b}^2}$ common, we get
${{\text{(asin}}\theta {\text{ + bcos}}\theta )^2}{\text{ = }}{{\text{a}}^2}{\text{(si}}{{\text{n}}^2}\theta {\text{ + co}}{{\text{s}}^2}\theta ){\text{ + }}{{\text{b}}^2}{\text{(co}}{{\text{s}}^2}\theta {\text{ + si}}{{\text{n}}^2}\theta ){\text{ - }}{{\text{c}}^2}$
As, ${\text{si}}{{\text{n}}^2}{\text{x + co}}{{\text{s}}^2}{\text{x = 1}}$,
Therefore,
${{\text{(asin}}\theta {\text{ + bcos}}\theta )^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - }}{{\text{c}}^2}$
Taking under – root both sides, we get
${\text{asin}}\theta {\text{ + bcos}}\theta {\text{ = }} \pm \sqrt {{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - }}{{\text{c}}^2}}$