NDA 2019 Paper 1 | Inverse Trigonometric Function Question 15 | Mathematics

If $${\sin ^{ - 1}}{{2p} \over {1 + {p^2}}} - {\cos ^{ - 1}}{{1 - {q^2}} \over {1 + {q^2}}} = {\tan ^{ - 1}}{{2x} \over {1 - {x^2}}}$$, then what is x equal to?

$${{p + q} \over {1 + pq}}$$

$${{p - q} \over {1 + pq}}$$

$${{pq} \over {1 + pq}}$$

$${{p + q} \over {1 - pq}}$$

NDA Mathematics 21 April 2024

MCQ (Single Correct Answer)

+2.5

-0.83

If $\cos^{-1} x = \sin^{-1} x$, then which one of the following is correct?

$x = 1$

$x = \frac{1}{2}$

$x = \frac{1}{\sqrt{2}}$

$x = \frac{1}{\sqrt{3}}$

NDA Mathematics 21 April 2024

MCQ (Single Correct Answer)

+2.5

-0.83

What is $\sqrt{15 + \cot^2 \left( \frac \pi 4 - 2 \cot^{-1} 3 \right)}$ equal to?

NDA Mathematics 21 April 2024

MCQ (Single Correct Answer)

+2.5

-0.83

What is $\tan^{-1} \left( \frac{a}{b} \right) - \tan^{-1} \left( \frac{a - b}{a + b} \right)$ equal to?

$-\frac{\pi}{4}$

$\frac{\pi}{4}$

$\tan^{-1} \left( \frac{a^2 - b^2}{a^2 + b^2} \right)$

$\tan^{-1} \left( \frac{2ab}{a^2 + b^2} \right)$

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