1) If 2 sin A = 1 = √2 cos B and 3π/2 < B < 2π, then find the value of tan A + tan B / cos A - cos B. 2) If sin A / 3 = sin B / 4 = 1/5 and A, B are angles in the second quadrant,... 1) If 2 sin A = 1 = √2 cos B and 3π/2 < B < 2π, then find the value of tan A + tan B / cos A - cos B. 2) If sin A / 3 = sin B / 4 = 1/5 and A, B are angles in the second quadrant, then prove that 4 cos A + 3 cos B = -5. 3) If tan θ = 2, evaluate 2 sin θ + 3 cos θ / 4 cos θ + 3 sin θ. 4) Eliminate θ from the following: i) x = 3 sec θ, y = 4 tan θ, ii) x = 6 cosec θ, y = 8 cot θ, iii) x = 4 cos θ - 5 sin θ, y = 4 sin θ + 5 cos θ, iv) x = 5 + 6 cosec θ, y = 3 + 8 cot θ, v) 2x = 3 - 4 tan θ, 3y = 5 + 3 sec θ. 5) If 2 sin² θ + 3 sin θ = 0, find the permissible values of cos θ. Read more

Steps to Solve

1. Use Given Trigonometric Identity

We start with the equation (2 \sin A = 1 = \sqrt{2} \cos B). This implies:

$$ \sin A = \frac{1}{2} $$ $$ \cos B = \frac{1}{\sqrt{2}} $$

2. Determine Angles A and B

From ( \sin A = \frac{1}{2} ),

• The angles in ( \left(\frac{\pi}{2}, \pi\right) ) satisfying this are ( A = \frac{\pi}{6}, \frac{5\pi}{6}. )

From ( \cos B = \frac{1}{\sqrt{2}} ),

• The angles in ( \left(\frac{3\pi}{2}, 2\pi\right) ) satisfying this are ( B = \frac{7\pi}{4}, \frac{5\pi}{4}. )

3. Apply The Given Range for B

The condition ( \frac{3\pi}{2} < B < 2\pi ) narrows it down:

• Therefore, ( B = \frac{7\pi}{4}. )

4. Calculate ( \tan A + \tan B )

We can find ( \tan A ) and ( \tan B ):

• For ( A = \frac{\pi}{6} ):

$$ \tan A = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} $$

• For ( B = \frac{7\pi}{4} ):

$$ \tan B = \tan\left(\frac{7\pi}{4}\right) = 1. $$

So,

$$ \tan A + \tan B = \frac{1}{\sqrt{3}} + 1. $$

5. Calculate ( \cos A - \cos B )

Next, find the cosines for ( A ) and ( B ):

• For ( A = \frac{\pi}{6} ):

$$ \cos A = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} $$

• For ( B = \frac{7\pi}{4} ):

$$ \cos B = \cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2}. $$

Then,

$$ \cos A - \cos B = \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}. $$

6. Calculate the Final Expression

Now, substitute into:

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

This becomes:

$$ \frac{\frac{1}{\sqrt{3}} + 1}{\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}}. $$

Now calculate the values.