Question

. If sin(a+B) = 1, sin(s–b) =1/2, a,b€[0,π/2] then find the value of tan(a +2B). tan (2a +B). 2​

Answer 1

Given,

$\blue{ { \:{ \boxed{\boxed{\begin{array}{cc} \rm \: sin( \alpha + \beta ) = 1 \\ \\ \rm \: \implies\: sin(\alpha + \beta) =sin {90}^{ \circ} \\ \\ \rm \implies\: \alpha + \beta = {90}^{ \circ} \: \: \: \: - - - - (1)\end{array}}}} }}$

and

$\pink{ \boxed{\boxed{\begin{array}{cc} \rm \: sin( \alpha - \beta ) = \frac{1}{2} \\ \\ \rm \implies\:sin( \alpha - \beta ) = sin {30}^{ \circ} \\ \\ \rm \implies\: \alpha - \beta = {30}^{ \circ} \: \: \: \: - - - - (2) \end{array}}}}$

where,

$\alpha , \beta \: \in [0, \frac{\pi}{2} ]$

Now,

$\orange{ \boxed{\boxed{\begin{array}{cc} \sf \: eqn(1) + eqn(2) \implies \\ \\ \rm \: \alpha + \beta = 90 \\ \alpha - \beta = 30 \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \rm \: 2 \alpha = 90 + 30 = 120 \\ \\ \boxed{\rm \therefore \: \alpha = {60}^{ \circ}} \\ \\ \sf \: put \: this \: value \: in \: eqn(1) \\ \\ \rm \: 60 + \beta = 90 \\ \\ \boxed{ \rm \therefore \: \beta = {30}^{ \circ}} \end{array}}}}$

Now, value of the following :

$\pink{ \boxed{\boxed{\begin{array}{cc} \rm \: tan( \alpha + 2 \beta ).tan(2 \alpha + \beta ) \\ \\ = \rm \: tan(60 + 2 \times 30).tan(2 \times 60 + 30) \\ \\ = \rm \: tan120.tan150 \\ \\ = \rm \: tan(90 + 30).tan(90 + 60) \\ \\ = \rm \: cot30.cot60 \\ \\ = \sqrt{3} \times \frac{1}{ \sqrt{3} } \\ \\ = 1 \end{array}}}}$