Question

if 2 sinA=1=√2 cosB and π/2<A<π , 3π/2<B<2π then find value of tanA+tanB/cosA-cosB​

Answer 1

Answer:

Given

2sinA = 1

⇒ sinA = 1/2 = sin30°

⇒ A = 30°

Also, √2 cosB = 1

⇒ cosB = 1/√2 = cos45°

⇒ B = 45°

Therefore

$\frac{\tan A+\tan B}{\cos A-\cos B}$

$=\frac{\tan 30^\circ+\tan 45^\circ}{\cos 30^\circ-\cos 45^\circ}$

$=\frac{(1/\sqrt{3}) +1}{(\sqrt{3}/2)-(1/\sqrt{2})}$

$=\frac{\sqrt{3}+1}{\sqrt{3}} \times \frac{2}{\sqrt{3}-\sqrt{2}}$

$=\frac{\sqrt{3}+3}{3} \times \frac{2(\sqrt{3}+\sqrt{2})}{3-2}$

$=\frac{2}{3}\times (\sqrt{3}+\sqrt{2})(\sqrt{3}+3})$

$=\frac{2}{3}(\sqrt{3}+\sqrt{2})(\sqrt{3}+3})$

Answer 2

Answer is give below in attachment

Step-by-step explanation: