Question

(cos+sin/cos-sin) - (cos-sin/cos+sin) = 4tan/1-tan2

Answer 1

Step-by-step explanation:

Taking LHS :-

• $( \frac{ \cos(a) + \sin(a) }{ \cos(a) - \sin(a) } ) - ( \frac{ \cos(a) - \sin(a) }{ \cos(a) + \sin(a) } )$
• $\frac{( \cos(a) + \sin(a) ) {}^{2} - ( \cos(a) - \sin(a) ) {}^{2} }{( \cos(a) - \sin(a))( \cos(a) + \sin(a) ) }$
• $because \: \: \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \: \: and \: {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \: \: \: \: \: and \: \: (a - b)(a + b) = {a}^{2} - {b}^{2}$
• Then, after solve this we get,
• $\frac{4 \sin(a) \cos(a) }{ \cos {}^{2} (a) \: - \sin {}^{2} (a) }$
• Now, because
• $2 \sin(a) \cos(a) = \sin(2a) \: \: and \: \cos {}^{2} (a) - \sin {}^{2} (a) = \cos(2a)$
• Then, we get
• $\frac{2 \sin(2a) }{ \cos(2a) }$
• $= > 2 \tan(2a)$
• But,
• $\tan(2a \: ) = \frac{2 \tan(a) }{1 - \tan {}^{2} (a) }$
• Then, we get : -
• $\frac{</strong><strong>4</strong><strong> \tan(a) }{1 - \tan {}^{2} (a) } \: = rhs$
• Hence proved.
• I hope it will help you. Thank you