Trigonometry practice dpp

Q1. In \(\triangle ABC\), \(\angle B = 90^\circ\). If \(AB = 14\) cm and \(AC = 50\) cm then \(\tan A\) equals:
Options: (A) \(\tfrac{24}{25}\) (B) \(\tfrac{24}{7}\) (C) \(\tfrac{7}{24}\) (D) \(\tfrac{25}{24}\)

Answer: (B) 24/7

Q2. If \(\sin\theta = \tfrac{12}{13}\) then the value of \(\displaystyle \frac{2\cos\theta + 3\tan\theta}{\sin\theta + \tan\theta\,\sin\theta}\) is:
Options: (A) \(\tfrac{12}{5}\) (B) \(\tfrac{5}{13}\) (C) \(\tfrac{259}{102}\) (D) \(\tfrac{259}{65}\)

Answer: (C) 259/102

Q3. If \(\sec\theta = \dfrac{\sqrt{p^2+q^2}}{q}\) then the value of \(\displaystyle \frac{p\sin\theta - q\cos\theta}{p\sin\theta + q\cos\theta}\) is:
Options: (A) \(\tfrac{p}{q}\) (B) \(\tfrac{p^2}{q^2}\) (C) \(\tfrac{p^2 - q^2}{p^2 + q^2}\) (D) \(\tfrac{p^2 + q^2}{p^2 - q^2}\)

Answer: (C) \(\tfrac{p^2 - q^2}{p^2 + q^2}\)

Q4. The value of \[ 2\tan^2 60^\circ \;-\;4\cos^2 45^\circ\;-\;3\sec^2 30^\circ \] is:
Options: (A) 0 (B) 1 (C) 12 (D) 8

Answer: (A) 0

Q5. The value of \[ \tfrac{3}{4}\tan^2 30^\circ \;-\;3\sin^2 60^\circ\;+\;\csc^2 45^\circ \] is:
Options: (A) 1 (B) 8 (C) 0 (D) 12

Answer: (C) 0

Q6. If \[ 7\sin^2\theta \;+\;3\cos^2\theta \;=\;4 \] then:
Options: (A) \(\tan\theta = \tfrac{1}{\sqrt2}\) (B) \(\tan\theta = \tfrac12\) (C) \(\tan\theta = \tfrac13\) (D) \(\tan\theta = \tfrac{1}{\sqrt3}\)

Answer: (D) \(\tan\theta = \tfrac{1}{\sqrt3}\)

Q7. The solution of the trigonometric equation $$ \frac{\cos^2\theta}{\cot^2\theta - \cos^2\theta} \;=\; 3, \quad 0^\circ<\theta<90^\circ $$
Options: (A) \(\theta = 15^\circ\) (B) \(\theta = 30^\circ\) (C) \(\theta = 60^\circ\) (D) \(\theta = 45^\circ\)

Answer: (C) \(\theta = 60^\circ\)

Q8. If \(\cot\theta + \cos\theta = p\) and \(\cot\theta - \cos\theta = q\), then the value of \(p^2 - q^2\) is:
Options: (A) \(2\sqrt{pq}\) (B) \(4\sqrt{pq}\) (C) \(2pq\) (D) \(4pq\)

Answer: (B) \(4\sqrt{pq}\)

Q9. If \(\alpha + \beta = 90^\circ\) and \(\alpha = 2\beta\) then \(\cos^2\alpha + \sin^2\beta\) equals:
Options: (A) 1 (B) 0 (C) \(\tfrac12\) (D) 2

Answer: (C) \(\tfrac12\)

Q10. If \(\cos(\alpha + \beta) = 0\), then \(\sin(\alpha - \beta)\) can be reduced to:
Options: (A) \(\cos\beta\) (B) \(\cos2\beta\) (C) \(\sin\alpha\) (D) \(\sin2\alpha\)

Answer: (B) \(\cos2\beta\)

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