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.

Question image

Understand the Problem

The question consists of multiple mathematical problems related to trigonometric identities and equations, requiring step-by-step solutions to find specific values or prove certain equations.

Answer

The simplified answer is $$ \frac{\frac{1}{\sqrt{3}} + 1}{\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}}. $$
Answer for screen readers

The final value is

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

This simplifies numerically as needed.

Steps to Solve

  1. Use Given Trigonometric Identity

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

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

  1. 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}. )
  1. Apply The Given Range for B

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

  • Therefore, ( B = \frac{7\pi}{4}. )
  1. 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,

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

  1. 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,

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

  1. 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.

The final value is

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

This simplifies numerically as needed.

More Information

This problem works through trigonometric identities and angle identities. It's important to use the correct quadrant in which angles lie based on given conditions.

Tips

  • Failing to determine the correct quadrant for angles A and B.
  • Not simplifying expressions correctly after substitution.
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