# If $a\cos \theta +b\sin \theta =m$ and $a\sin \theta -b\cos \theta =n,$ then the value of ${{a}^{2}}+{{b}^{2}}$ is

a) $m+n$

b) $mn$

c) $\sqrt{mn}$

d) ${{m}^{2}}+{{n}^{2}}$

Last updated date: 20th Mar 2023

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Hint: To solve this question, we will be using the trigonometric identity given by${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$. We have two equations given in the question, we have to take the square of both the equations and then add them to get the answer.

Complete step-by-step answer:

Before moving to the solution we have to be very clear about the rules of each trigonometric ratios, the basic formula related to trigonometric identities, the maximum and minimum values of trigonometric ratios like the maximum value of $\sin \theta =1$, $cos\theta =1$, $tan\theta =\infty $ and minimum values of $\sin \theta =1$, $cos\theta =1$, $tan\theta =\infty $ , ranges of $\sin \theta \in \left( -1,1 \right)$, $\cos \theta \in \left( -1,1 \right)$, $\tan \theta \in \left( -\infty ,\infty \right)$ trigonometric ratios.

The equations we have:

$\begin{align}

& a\cos \theta +b\sin \theta =m......(i) \\

& a\sin \theta -b\cos \theta =n.......(ii) \\

\end{align}$

Squaring the equations (i) and (ii) on both the sides we get,

${{(a\cos \theta +b\sin \theta )}^{2}}={{m}^{2}}.....(iii)$

${{(a\sin \theta -b\cos \theta )}^{2}}={{n}^{2}}.....(iv)$

Adding equations (iii) and (iv) we get,

$\Rightarrow {{\left( a\cos \theta +b\sin \theta \right)}^{2}}+{{\left( a\sin \theta -b\cos \theta \right)}^{2}}={{m}^{2}}+{{n}^{2}}$

$\Rightarrow ({{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +2ab\cos \theta \sin \theta )+({{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2ab\cos \theta \sin \theta )={{m}^{2}}+{{n}^{2}}$

$\Rightarrow {{a}^{2}}{{\sin }^{2}}\theta +{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta -2ab\cos \theta +2ab\cos \theta ={{m}^{2}}+{{n}^{2}}$ $\Rightarrow \left( {{a}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+{{b}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+2ab\sin \theta \cos \theta -2ab\cos \theta sin\theta \right)={{m}^{2}}+{{n}^{2}}$

Using ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ we get,

${{a}^{2}}+{{b}^{2}}={{m}^{2}}+{{n}^{2}}$

Therefore, the value of ${{a}^{2}}+{{b}^{2}}={{m}^{2}}+{{n}^{2}}$.

Hence, the answer is option (d).

Note: We have alternate method to solve this question is about finding the value of${{a}^{2}}+{{b}^{2}}$so, from the question it is very clear that we will get the answer only by squaring both the equations together, so if you are asked for the answer from the options, just square and add the right-hand side of both the equations. You will get your answer as option (d).

There is a possibility that after looking at the question, students generally use to multiply both the equations together in that case, we will get the value of ${{a}^{2}}-{{b}^{2}}$ not ${{a}^{2}}+{{b}^{2}}$ so be careful about the question. While squaring both sides in the case of quadratic equations, be careful about the roots because we may lose extra roots during cancellation.

Complete step-by-step answer:

Before moving to the solution we have to be very clear about the rules of each trigonometric ratios, the basic formula related to trigonometric identities, the maximum and minimum values of trigonometric ratios like the maximum value of $\sin \theta =1$, $cos\theta =1$, $tan\theta =\infty $ and minimum values of $\sin \theta =1$, $cos\theta =1$, $tan\theta =\infty $ , ranges of $\sin \theta \in \left( -1,1 \right)$, $\cos \theta \in \left( -1,1 \right)$, $\tan \theta \in \left( -\infty ,\infty \right)$ trigonometric ratios.

The equations we have:

$\begin{align}

& a\cos \theta +b\sin \theta =m......(i) \\

& a\sin \theta -b\cos \theta =n.......(ii) \\

\end{align}$

Squaring the equations (i) and (ii) on both the sides we get,

${{(a\cos \theta +b\sin \theta )}^{2}}={{m}^{2}}.....(iii)$

${{(a\sin \theta -b\cos \theta )}^{2}}={{n}^{2}}.....(iv)$

Adding equations (iii) and (iv) we get,

$\Rightarrow {{\left( a\cos \theta +b\sin \theta \right)}^{2}}+{{\left( a\sin \theta -b\cos \theta \right)}^{2}}={{m}^{2}}+{{n}^{2}}$

$\Rightarrow ({{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +2ab\cos \theta \sin \theta )+({{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2ab\cos \theta \sin \theta )={{m}^{2}}+{{n}^{2}}$

$\Rightarrow {{a}^{2}}{{\sin }^{2}}\theta +{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta -2ab\cos \theta +2ab\cos \theta ={{m}^{2}}+{{n}^{2}}$ $\Rightarrow \left( {{a}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+{{b}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+2ab\sin \theta \cos \theta -2ab\cos \theta sin\theta \right)={{m}^{2}}+{{n}^{2}}$

Using ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ we get,

${{a}^{2}}+{{b}^{2}}={{m}^{2}}+{{n}^{2}}$

Therefore, the value of ${{a}^{2}}+{{b}^{2}}={{m}^{2}}+{{n}^{2}}$.

Hence, the answer is option (d).

Note: We have alternate method to solve this question is about finding the value of${{a}^{2}}+{{b}^{2}}$so, from the question it is very clear that we will get the answer only by squaring both the equations together, so if you are asked for the answer from the options, just square and add the right-hand side of both the equations. You will get your answer as option (d).

There is a possibility that after looking at the question, students generally use to multiply both the equations together in that case, we will get the value of ${{a}^{2}}-{{b}^{2}}$ not ${{a}^{2}}+{{b}^{2}}$ so be careful about the question. While squaring both sides in the case of quadratic equations, be careful about the roots because we may lose extra roots during cancellation.

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