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

Answer 1

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.

Complete step-by-step answer:
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}} $
So, answer is option (B)

Note: To solve such types of questions in which a condition is given and we have to find the value of an expression, we will use a technique to solve the question without any error. Firstly, we will see which property we have to apply for. Such questions at first look difficult, when the student does not know how to solve them, but in reality, such questions are very easy. We just have to find the value from the given condition and put in the expression to get the answer. After the first step, we have to perform mathematical operations (multiplication, division, etc) based on given options to find the value of a variable. After it, we will apply the same operation on the expression whose value we have to find and put the value of the variable in the expression to get the answer.