JEE Main Previous Year Trigonometry Questions and Solutions - Solution of Triangles Trigonometric Identities, Trigonometry

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Solution

Q.
If $\tan A = \frac{1}{\sqrt{x (x^{2} + x + 1)}},$ $\tan B = \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}$ and $\tan C = {(x^{- 3} + x^{- 2} + x^{- 1})}^{1 / 2}, 0 < A, B, C < π / 2,$ then $A + B$ is equal to: [2024]

1 $π - C$

2 $2 π - C$

3 $C$

4 $\frac{π}{2} - C$

Ans.
(3)

As $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$= \frac{\frac{1}{\sqrt{x (x^{2} + x + 1)}} + \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}}{1 - \frac{1}{x^{2} + x + 1}}$

$= \frac{1 + x}{\sqrt{x (x^{2} + x + 1)}} \times \frac{x^{2} + x + 1}{x^{2} + x} = \frac{1 + x}{\sqrt{x}} \times \frac{\sqrt{x^{2} + x + 1}}{x (1 + x)}$

$= \sqrt{\frac{x^{2} + x + 1}{x^{3}}} = \sqrt{x^{- 1} + x^{- 2} + x^{- 3}} = \tan C$

$\Rightarrow A + B = C$

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