Question

$\begin{array}{l}\operatorname{Tan}(a-b)=1 \ \sec(a+b)=2 / \sqrt{3} \ \text{Find } a, b\end{array}$

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Answer

The problem statement gives us two trigonometric equations:

  1. $$ \tan(a-b) = 1 $$
  2. $$ \sec(a+b) = \frac{2}{\sqrt{3}} $$

We are asked to find the values for $a$ and $b$ given these conditions.

Step-by-Step Analysis

Equation 1 Analysis: The equation $\tan(a-b) = 1$ usually implies that $a-b$ could be equal to an angle where the tangent value is 1. The general solutions for this scenario are: $$ a - b = \frac{\pi}{4} + n\pi $$ where $n$ is any integer.

Equation 2 Analysis: For the equation $\sec(a+b) = \frac{2}{\sqrt{3}}$, knowing values where $\cos$ is $\frac{\sqrt{3}}{2}$ is useful because $\sec$ is the reciprocal of $\cos$. This usually happens at: $$ a + b = \pm\frac{\pi}{6} + 2n\pi $$ Here again, $n$ is any integer.

Solving for $a$ and $b$

Given the periodicity of trigonometric functions and the nature of these specific values, the most straightforward choice for $n$ would be $0$ to simplify calculations.

Thus, we have: $$ a - b = \frac{\pi}{4} $$ and $$ a + b = \frac{\pi}{6} $$

Adding these two equations, we get: $$ 2a = \frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi + 2\pi}{12} = \frac{5\pi}{12} $$ $$ a = \frac{5\pi}{24} $$

Subtracting the $a - b$ equation from the $a + b$ equation: $$ 2b = \frac{\pi}{6} - \frac{\pi}{4} = \frac{2\pi - 3\pi}{12} = -\frac{\pi}{12} $$ $$ b = -\frac{\pi}{24} $$

Hence, the values of $a$ and $b$ that satisfy both conditions are: $$ a = \frac{5\pi}{24},, b = -\frac{\pi}{24} $$

These values can of course vary by adding integer multiples of $\pi$ or $2\pi$, depending on the periodicity of the trigonometric functions involved.


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