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## QuestionMathsClass 10

If a and b are real number such that $$acos\theta +bsin\theta =4$$ and $$asin\theta -bcos\theta =3$$, then $${(a}^{2}+{b}^{2})$$ is
(A) 7
(B) 12
(C) 25
(D) $$\sqrt{12}$$

(C)
4.6
4.6

## Solution

$$acos\theta +bsin\theta =4$$ ______ (1)
$$asin\theta -bcos\theta =3$$ _________(2)
$$(1{)}^{2}+(2{)}^{2}$$
$$=(acos\theta +bsin\theta {)}^{2}(asin\theta -bcos\theta {)}^{2}={4}^{2}+{3}^{2}$$
$${a}^{2}co{s}^{2}\theta +{b}^{2}si{n}^{2}\theta +ab\;sin\theta\;cos\theta +{a}^{2}si{n}^{2}\theta +{b}^{2}co{s}^{2}\theta -2ab\;sin\theta\;cos\theta =16+9$$
$${a}^{2}(co{s}^{2}\theta +si{n}^{2}\theta )+{b}^{2}(si{n}^{2}\theta +co{s}^{2}\theta )=25$$
$${a}^{2}+{b}^{2}=25$$
$$(\therefore\;si{n}^{2}\theta +co{s}^{2}\theta =1$$