Question

Column - 1: Trigonometric expression
Column - 2: Dependency on '$\alpha$' and '$\beta$'
Column - 3: Equivalent value of Trigonometric expression in Column - 1.

$\begin{array}{cccccc}& Column-1& & Column-2& & Column-3\\ \left(I\right)& {\cos }^2(\alpha +\beta )-{\sin }^2(\alpha -\beta )-\cos \left(2\alpha \right)\cos \left(2\beta \right)+1& \left(i\right)& \text{Independent of}\alpha \text{only}& \left(P\right)& 1\\ \left(II\right)& {\cos }^2\alpha +{\cos }^2(\alpha +\beta )-2\cos \left(\alpha \right)\cos \left(\beta \right)\cos (\alpha +\beta )& \left(ii\right)& \text{Independent of}\beta \text{only}& \left(Q\right)& {\sin }^2\beta \\ \left(III\right)& \frac{\sin (\alpha -\beta )\sin (\alpha +\beta )}{1-{\tan }^2\alpha {\cot }^2\beta }& \left(iii\right)& \text{Independent of}\alpha \text{and}\beta & \left(R\right)& -{\cos }^2\alpha {\sin }^2\beta \\ \left(IV\right)& 2{\sin }^2\beta +4\cos (\alpha +\beta )\sin \alpha \sin \beta +\cos 2(\alpha +\beta )& & & \left(S\right)& \cos \left(2\alpha \right)\end{array}$
Which of the following options is the only correct combination ?

• A. (IV)(i)(Q)
• B. (IV)(ii)(S)
• C. (III)(ii)(S)
• D. (III)(i)(Q)

Answer 1

The correct option is B (IV)(ii)(S)

(A) $\left(I\right)1+{\cos }^2(\alpha +\beta )-{\sin }^2(\alpha -\beta )-\cos \left(2\alpha \right)\cos \left(2\beta \right)=1+\cos 2\alpha \cos 2\beta -\cos 2\alpha \cos 2\beta =1$

$\left(II\right){\cos }^2\alpha +{\cos }^2(\alpha +\beta )-\cos (\alpha +\beta )\left[\cos \right(\alpha +\beta )+\cos (\alpha -\beta \left)\right]={\cos }^2\alpha -[{\cos }^2\alpha -{\sin }^2\beta ]={\sin }^2\beta$

$\left(III\right)\frac{({\sin }^2\alpha -{\sin }^2\beta ){\cos }^2\alpha {\sin }^2\beta }{{\cos }^2\alpha {\sin }^2\beta -{\sin }^2\alpha {\cos }^2\beta }$

$=\frac{({\sin }^2\alpha -{\sin }^2\beta ){\cos }^2\alpha {\sin }^2\beta }{\sin (\alpha +\beta )\sin (\beta -\alpha )}$

$=-{\cos }^2\alpha {\sin }^2\beta$

$\left(IV\right)2{\sin }^2\beta +2\cos (\alpha +\beta )\left[\cos \right(\alpha -\beta )-\cos (\alpha +\beta \left)\right]+2{\cos }^2(\alpha +\beta )-1$

$=2\left({\sin }^2\beta \right)-12{\cos }^2(\alpha +\beta )+2{\cos }^2(\alpha +\beta )+2({\cos }^2\alpha -{\sin }^2\beta )$

$=2{\cos }^2\alpha -1=\cos 2\alpha$