Question

Suppose alpha and beta are greater than zero. Alpha +2 beta is equal to $\frac{\pi }{2}$ then tan ($\alpha +\beta$) -2tan$\alpha$ -tan$\beta$ is equal to

Answer 1

Step-by-step explanation:

We have,

$\alpha + 2 \beta = \frac{\pi}{2}$

$\implies\alpha + \beta = \frac{\pi}{2} - \beta$

$\implies \tan( \alpha + \beta) = \tan \bigg( \frac{\pi}{2} - \beta \bigg)$

$\implies \tan( \alpha + \beta) = \cot( \beta )$

Now,

$\implies \tan( \alpha + \beta) - 2 \tan( \alpha ) - \tan( \beta ) = \cot( \beta ) - 2 \tan( \alpha ) - \tan( \beta )$

$\implies \tan( \alpha + \beta) - 2 \tan( \alpha ) - \tan( \beta ) = \cot \bigg( \frac{\pi}{4} - \frac{ \alpha }{2} \bigg) - 2 \tan( \alpha ) - \tan \bigg( \frac{\pi}{4} - \frac{ \alpha }{2} \bigg)$

$\implies \tan( \alpha + \beta) - 2 \tan( \alpha ) - \tan( \beta ) = \frac{1 + \tan( \frac{ \alpha }{2} ) }{1 - \tan( \frac{ \alpha }{2} ) } - 2 \tan( \alpha ) - \frac{1 - \tan( \frac{ \alpha }{2} ) }{1 + \tan( \frac{ \alpha }{2} ) }$

$\implies \tan( \alpha + \beta) - 2 \tan( \alpha ) - \tan( \beta ) = \frac{1 + \tan( \frac{ \alpha }{2} ) }{1 - \tan( \frac{ \alpha }{2} ) } - \frac{1 - \tan( \frac{ \alpha }{2} ) }{1 + \tan( \frac{ \alpha }{2} ) } - 2 \tan( \alpha )$

$\implies \tan( \alpha + \beta) - 2 \tan( \alpha ) - \tan( \beta ) = \frac{ \{1 + \tan( \frac{ \alpha }{2} ) \}^{2} - \{1 - \tan( \frac{ \alpha }{2} ) \}^{2} }{ \{1 - \tan( \frac{ \alpha }{2} ) \} \{ 1 + \tan( \frac{ \alpha }{2} ) \} } - 2 \tan( \alpha )$

$\implies \tan( \alpha + \beta) - 2 \tan( \alpha ) - \tan( \beta ) = \frac{ 4 \tan( \frac{ \alpha }{2} ) }{ 1 - \tan^{2} ( \frac{ \alpha }{2} )} - 2 \tan( \alpha )$

$\implies \tan( \alpha + \beta) - 2 \tan( \alpha ) - \tan( \beta ) =2. \frac{ 2 \tan( \frac{ \alpha }{2} ) }{ 1 - \tan^{2} ( \frac{ \alpha }{2} )} - 2 \tan( \alpha )$

$\implies \tan( \alpha + \beta) - 2 \tan( \alpha ) - \tan( \beta ) =2 \tan( \alpha ) - 2 \tan( \alpha )$

$\tt \purple{\implies \tan( \alpha + \beta) - 2 \tan( \alpha ) - \tan( \beta ) =0}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: \alpha , \beta > 0$

and

$\rm :\longmapsto\: \alpha + 2 \beta = \dfrac{\pi}{2}$

$\bf\implies \: \alpha + \beta = \dfrac{\pi}{2} - \beta - - - (1)$

Now, Consider,

$\rm :\longmapsto\:tan( \alpha + \beta) \: - 2tan \alpha - tan\beta$

$\rm \:  =  \: tan\bigg[\dfrac{\pi}{2} - \beta \bigg] - 2tan\alpha - tan\beta$

$\rm \:  =  \:cot\beta- 2tan\alpha - tan\beta$

$\rm \:  =  \:cot\beta - tan\beta - 2tan\alpha$

$\rm \:  =  \:\dfrac{1}{tan\beta } - tan\beta - 2tan\alpha$

$\rm \:  =  \:\dfrac{1 - {tan}^{2}\beta }{tan\beta } - 2tan\alpha$

$\rm \:  =  \dfrac{1}{\dfrac{tan\beta }{1 - {tan}^{2} \beta } } - 2tan\alpha$

can be rewritten as

$\rm \:  =  \dfrac{2}{\dfrac{2tan\beta }{1 - {tan}^{2} \beta } } - 2tan\alpha$

$\rm \:  =  \dfrac{2}{tan2\beta } - 2tan\alpha$

$\rm \:  =  \: 2cot2\beta - 2tan\alpha$

$\rm \:  =  \: 2cot\bigg[\dfrac{\pi}{2} - \alpha \bigg] - 2tan\alpha$

$\rm \:  =  \: 2tan\alpha - 2tan\alpha$

$\rm \:  =  \: 0$

Hence,

$\red{\rm :\longmapsto\:\red{ \boxed{ \sf{ \:tan( \alpha + \beta) \: - 2tan \alpha - tan\beta = 0}}}}$

• So, option (a) is correct

Formula Used :-

$\red{ \boxed{ \sf{ \:tan\bigg[\dfrac{\pi}{2} - x \bigg] = cotx}}}$

$\red{ \boxed{ \sf{ \:cot\bigg[\dfrac{\pi}{2} - x \bigg] = tanx}}}$

$\green{ \boxed{ \sf{ \:tan2x = \frac{2tanx}{1 - {tan}^{2} x} }}}$