Question
In the following match each item given under the $\text{column}\  C_1$ to its correct answer given under the $\text{column}\  C_2$:
  $\text{column}\  C_1$   $\text{column}\  C_2$
$(a)$ $\sin(\text{x + y})\sin\text{x}-\text{y}$ $(i)$ $\cos^2\text{x}-\sin^2\text{y}$
$(b)$ $\cos(\text{x + y})\cos(\text{x}-\text{y})$ $(ii)$ $\frac{1-\tan\theta}{1+\tan\theta}$
$(c)$ $\cot\Big(\frac{\pi}{4}+\theta\Big)$ $(iii)$ $\frac{1+\tan\theta}{1-\tan\theta}$
$(d)$ $\tan\Big(\frac{\pi}{4}+\theta\Big)$ $(iv)$ $\sin^2\text{x}-\sin^2\text{y}$

Answer

  $\text{column}\  C_1$   $\text{column}\  C_2$
$(a)$ $\sin(\text{x + y})\sin\text{x}-\text{y}$ $(iv)$ $\sin^2\text{x}-\sin^2\text{y}$
$(b)$ $\cos(\text{x + y})\cos(\text{x}-\text{y})$ $(i)$ $\cos^2\text{x}-\sin^2\text{y}$
$(c)$ $\cot\Big(\frac{\pi}{4}+\theta\Big)$ $(ii)$ $\frac{1-\tan\theta}{1+\tan\theta}$
$(d)$ $\tan\Big(\frac{\pi}{4}+\theta\Big)$ $(ii)$ $\frac{1+\tan\theta}{1-\tan\theta}$
$\sin(\text{x + y})\sin(\text{x}-\text{y})=\sin^2\text{x}-\sin^2\text{y}$
$\cos(\text{x + y})\cos(\text{x}-\text{y})\cos^2\text{x}-\cos^2\text{y}$
$\cot\Big(\frac{\pi}{4}+\theta\Big)=\frac{\frac{\cot\pi}{4}\cot\theta-1}{\cot\theta+\cot\frac{\pi}{4}}$
$=\frac{\cot\theta-1}{\cot\theta+1}=\frac{1-\tan\theta}{1+\tan\theta}$
$\tan\Big(\frac{\pi}{4}+\theta\Big)=\frac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\tan\theta}=\frac{1+\tan\theta}{1-\tan\theta}$

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