Question

Prove that

\[\left\{ 1 + \cot x - \sec\left( \frac{\pi}{2} + x \right) \right\}\left\{ 1 + \cot x + \sec\left( \frac{\pi}{2} + x \right) \right\} = 2\cot x\]

 

Answer 1

LHS =\[ \left\{ 1 + \cot x - \sec\left( \frac{\pi}{2} + x \right) \right\}\left\{ 1 + \cot x + \sec\left( \frac{\pi}{2} + x \right) \right\}\]
\[ = \left[ 1 + \cot x - \left\{ - cosec x \right\} \right]\left[ 1 + \cot x + \left\{ - cosec x \right\} \right] \]
\[ = \left[ 1 + \cot x + cosec x \right] \left[ 1 + \cot x - cosec x \right]\]
\[ = \left[ 1 + \cot x + cosec x \right] \left[ 1 + \cot x - cosec x \right]\]
\[ = \left[ \left\{ 1 + \cot\left( x \right) \right\} + \left\{ cosec x \right\} \right] \left[ \left\{ 1 + \cot x \right\} - \left\{ cosec x \right\} \right]\]
\[ = \left\{ 1 + \cot x \right\}^2 - \left\{ cosec x \right\}^2 \]
\[= 1 + \cot^2 x + 2\cot x - {cosec}^2 x\]
\[ = 2 \cot x \left[ \because 1 + \cot^2 x = {cosec}^2 x \right]\]
 = RHS 
Hence proved.