JEE 2022 — Mathematics PYQ

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Question

JEE 2022 — Mathematics PYQ

JEE | Mathematics | 2022

The value of the expression $50 tan (3 tan^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{5})) + 42 tan (\frac{1}{2} tan^{- 1} (22))$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1: Simplify the first part of the tan argument.

Let $\alpha = \tan^{-1} \left( \frac{1}{2} \right)$ and $\beta = \cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$ .

For $\beta$ , if $\cos \beta = \frac{1}{\sqrt{5}}$ , then using a right triangle, the opposite side is $\sqrt{(\sqrt{5})^2 - 1^2} = 2$ .

So, $\tan \beta = \frac{2}{1} = 2 \implies \beta = \tan^{-1}(2)$ .

The argument is $3\alpha + 2\beta$ . Note that $\tan \alpha = \frac{1}{2}$ and $\tan \beta = 2$ , which means $\alpha + \beta = \frac{\pi}{2}$ .

Substituting $\beta = \frac{\pi}{2} - \alpha$ :

3\alpha + 2\left( \frac{\pi}{2} - \alpha \right) = 3\alpha + \pi - 2\alpha = \pi + \alpha

Thus, $\tan(\pi + \alpha) = \tan \alpha = \frac{1}{2}$ .

The first term becomes: $50 \times \frac{1}{2} = 25$ .

Step 2: Simplify the second term.

Let $\phi = \tan^{-1}(2\sqrt{2})$ , so $\tan \phi = 2\sqrt{2}$ . We need to find $\tan \left( \frac{\phi}{2} \right)$ .

Using the identity $\tan \phi = \frac{2\tan(\phi/2)}{1 - \tan^2(\phi/2)}$ :

2\sqrt{2} = \frac{2t}{1 - t^2} \implies \sqrt{2} - \sqrt{2}t^2 = t \implies \sqrt{2}t^2 + t - \sqrt{2} = 0

Solving using the quadratic formula:

t = \frac{-1 \pm \sqrt{1 - 4(\sqrt{2})(-\sqrt{2})}}{2\sqrt{2}} = \frac{-1 \pm \sqrt{1 + 8}}{2\sqrt{2}} = \frac{-1 \pm 3}{2\sqrt{2}}

Since $\phi$ is in the first quadrant, $t$ must be positive:

t = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}

The second term becomes: $4\sqrt{2} \times \frac{1}{\sqrt{2}} = 4$ .

Step 3: Final Addition

\text{Total} = 25 + 4 = 29

Answer: The value is 29.