Question

If $\cos (\theta -\alpha )=a,\sin (\theta -\beta )=b$, then $a^2-2ab\sin (\alpha -\beta )+b^2$ is equal to

• A. ${\sin }^2(\alpha –\beta )$
• B. ${\cos }^2(\alpha –\beta )$
• C. $1$
• D. $0$

Answer 1

The correct option is B

${\cos }^2(\alpha –\beta )$

Step 1. Find the value of ${\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathit{a}\mathit{b}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{\alpha }\mathbf{}\mathbf{-}\mathbf{}\mathit{\beta }\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}{\mathit{b}}^{\mathbf{2}}$:

Given, $\cos (\theta -\alpha )=a,\sin (\theta -\beta )=b$

$\Rightarrow$$\alpha –\beta =(\theta –\beta )–(\theta –\alpha )$

Step 2. Take “sin” on both sides,

$\begin{array}{rcl}\sin (\alpha –\beta )& =& \sin \left[\right(\theta –\beta )–(\theta –\alpha \left)\right]\\ & =& \sin (\theta –\beta )\cos (\theta –\alpha )–\cos (\theta –\beta )\sin (\theta –\alpha )\\ & =& ba–\sqrt{(1–b^2)}\sqrt{(1–a^2)}\end{array}$

$\sin (\alpha -\beta )=ab–\sqrt{(1–a^2)}\sqrt{(1–b^2)}$ ….(i)

Again,

$\alpha –\beta =(\theta –\beta )–(\theta –\alpha )$

Step 3. Take “cos” on both sides,

$\begin{array}{rcl}\cos (\alpha –\beta )& =& \cos \left[\right(\theta –\beta )–(\theta –\alpha \left)\right]\\ & =& \cos (\theta –\alpha )\cos (\theta –\beta )+\sin (\theta –\alpha )\sin (\theta –\beta )\\ & =& a\sqrt{(1–b^2)}+b\sqrt{(1–a^2)}\end{array}$

${\cos }^2(\alpha –\beta )=a^2(1–b^2)+b^2(1–a^2)+2ab\sqrt{(1–a^2)}\sqrt{(1–b^2)}$ ….(ii)

$\therefore a^2–2ab\sin (\alpha –\beta )+b^2=a^2–2ab[ab–\sqrt{(1–a^2)}\sqrt{(1–b^2)}]+b^2$ {from (i)}

$=a^2–2a^2{b}^2+2ab\sqrt{(1–a^2)}\sqrt{(1–b^2)}+b^2$ …..(iii)

Step 4. By equating equation (ii) and (iii), we get

${\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}\mathbf{2}\mathit{a}\mathit{b}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{\alpha }\mathbf{}\mathbf{–}\mathbf{}\mathit{\beta }\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}{\mathit{b}}^2\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{(}\mathit{\alpha }\mathbf{}\mathbf{–}\mathbf{}\mathit{\beta }\mathbf{)}$

Hence, Option ‘B’ is Correct.