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  • In the following match each item given under the column C1 to its correct answer given under the column C2 a):sin(x+y)sin(x-y)i)cos^2x-sin^2y

In the following match each item given under the column C_1 to its correct answer given under the column C_2:

\text { (a) } \sin (\mathrm{x}+\mathrm{y}) \sin (\mathrm{x}-\mathrm{y}) \text { (i) } \cos ^{2} \mathrm{x}-\sin ^{2} \mathrm{y}
\text { (b) } \cos (\mathrm{x}+\mathrm{y}) \cos (\mathrm{x}-\mathrm{y}) \text { (ii) } \frac{1-\tan \theta}{1+\tan \theta}
\text { (c) } \cot \left(\frac{\pi}{4}+\theta\right) \text { (iii) } \frac{1+\tan \theta}{1-\tan \theta}
\text { (d) } \tan \left(\frac{\pi}{4}+\theta\right) \text { (iv) } \sin ^{2} \mathrm{x}-\sin ^{2} \mathrm{y}

 

Answers (1)

\\ \sin \left( x+y \right) \sin \left( x - y \right) =\sin ^{2}x - \sin ^{2}y \\\\ \cos \left( x+y \right) \cos \left( x - y \right) =\cos ^{2}x - \cos ^{2}y \\\\ \cot \left( \frac{ \pi }{4}+ \theta \right) =\frac{\cot \frac{ \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 \left( \frac{ \pi }{4}+ \theta \right) =\frac{\tan \frac{ \pi }{4}\tan \theta - 1}{\tan \theta +\tan \frac{ \pi }{4}}=\frac{1+\tan \theta }{1 - \tan \theta } \\\\

Thus, (a) - (iv) , (b) - (i), (c) -(ii), (d) - (iii) 

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