Question

If 2 sinA = 1 = √2 cosB and π
/2 < A < π,

3π
/2< B < 2π, then find the value of

tanA+tanB/cosA-cosB​

Answer 1

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

Given that

$\rm :\longmapsto\:2sinA = 1 \: \: and \: \dfrac{\pi}{2} < A < \pi$

$\rm :\longmapsto\:sinA = \dfrac{1}{2}$

As A lies in second quadrant

$\rm :\longmapsto\:A = \dfrac{5\pi}{6}$

Consider,

$\rm :\longmapsto\:cosA$

$\rm \: \: = \: cos\dfrac{5\pi}{6}$

$\rm \: \: = \: cos \bigg(\pi - \dfrac{\pi}{6} \bigg)$

$\rm \: \: = - \: cos \bigg(\dfrac{\pi}{6} \bigg)$

$\rm \: \: = \: - \: \dfrac{ \sqrt{3} }{2}$

$\bf\implies \:cosA = - \: \dfrac{ \sqrt{3} }{2}$

Consider,

$\rm :\longmapsto\:tanA$

$\rm \: \: = \: tan\dfrac{5\pi}{6}$

$\rm \: \: = \: tan \bigg(\pi - \dfrac{\pi}{6} \bigg)$

$\rm \: \: = - \: tan \bigg(\dfrac{\pi}{6} \bigg)$

$\rm \: \: = \: - \: \dfrac{1}{ \sqrt{3} }$

$\bf\implies \:tanA = - \: \dfrac{1}{ \sqrt{3} }$

Also,

Given that,

$\rm :\longmapsto\: \sqrt{2}cosB = 1 \: \: and \: \dfrac{3\pi}{2} < B < 2\pi$

$\rm :\longmapsto\:cosB = \dfrac{1}{ \sqrt{2} }$

As B lies in fourth quadrant,

$\rm :\longmapsto\:B = \dfrac{7\pi}{4}$

Consider,

$\rm :\longmapsto\:tanB$

$\rm \: \: = \: tan\dfrac{7\pi}{4}$

$\rm \: \: = \: tan \bigg(2\pi - \dfrac{\pi}{4} \bigg)$

$\rm \: \: = - \: tan \bigg(\dfrac{\pi}{4} \bigg)$

$\rm \: \: = \: - \: 1$

$\bf\implies \:tanB = - \: 1$

Consider,

$\bf :\longmapsto\:\dfrac{tanA + tanB}{cosA - cosB}$

$\rm \: \: = \: \dfrac{ - \dfrac{1}{ \sqrt{3} } \: - \: 1}{ - \: \dfrac{ \sqrt{3} }{2} \: - \: \dfrac{1}{ \sqrt{2} } }$

$\rm \: \: = \: \dfrac{ \dfrac{ - 1 - \sqrt{3} }{ \sqrt{3} } \: }{ \: \dfrac{ - \sqrt{3} - \sqrt{2} }{2} }$

$\rm \: \: = \: \dfrac{ \dfrac{ 1 + \sqrt{3} }{ \sqrt{3} } \: }{ \: \dfrac{\sqrt{3} + \sqrt{2} }{2} }$

$\rm \: \: = \: \dfrac{2 + 2 \sqrt{3} }{3 + \sqrt{6} }$

$\rm \: \: = \: \dfrac{2 + 2 \sqrt{3} }{3 + \sqrt{6} } \times \dfrac{3 - \sqrt{6} }{3 - \sqrt{6} }$

$\rm \: \: = \: \dfrac{6 - 2 \sqrt{6} + 6 \sqrt{3} - 6 \sqrt{2} }{9 - 6}$

$\rm \: \: = \: \dfrac{6 - 2 \sqrt{6} + 6 \sqrt{3} - 6 \sqrt{2} }{3}$

Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Sign of Trigonometric ratios in Quadrants

sin (90°-θ)  =  cos θ

cos (90°-θ)  =  sin θ

tan (90°-θ)  =  cot θ

csc (90°-θ)  =  sec θ

sec (90°-θ)  =  csc θ

cot (90°-θ)  =  tan θ

sin (90°+θ)  =  cos θ

cos (90°+θ)  =  -sin θ

tan (90°+θ)  =  -cot θ

csc (90°+θ)  =  sec θ

sec (90°+θ)  =  -csc θ

cot (90°+θ)  =  -tan θ

sin (180°-θ)  =  sin θ

cos (180°-θ)  =  -cos θ

tan (180°-θ)  =  -tan θ

csc (180°-θ)  =  csc θ

sec (180°-θ)  =  -sec θ

cot (180°-θ)  =  -cot θ

sin (180°+θ)  =  -sin θ

cos (180°+θ)  =  -cos θ

tan (180°+θ)  =  tan θ

csc (180°+θ)  =  -csc θ

sec (180°+θ)  =  -sec θ

cot (180°+θ)  =  cot θ

sin (270°-θ)  =  -cos θ

cos (270°-θ)  =  -sin θ

tan (270°-θ)  =  cot θ

csc (270°-θ)  =  -sec θ

sec (270°-θ)  =  -csc θ

cot (270°-θ)  =  tan θ

sin (270°+θ)  =  -cos θ

cos (270°+θ)  =  sin θ

tan (270°+θ)  =  -cot θ

csc (270°+θ)  =  -sec θ

sec (270°+θ)  =  cos θ

cot (270°+θ)  =  -tan θ