CBSE 10 Mathematics Trigonometry Practice Paper 08
1. Given \(tan A = \frac{4} {3}\), find the other trigonometric ratios of the angle A.
2. If \(\angle B\) and \(\angle Q\) are acute angles such that \(sin B = sin Q\), then prove that \(\angle B = \angle Q\).
3. Given \(15 cot A = 8\), find \(sin A\) and \(sec A\).
4. If \(3cot A = 4\), check whether \(\frac{1 - tan^2 A }{1 + tan^2 A }\) =\( cos^2 A - sin^2A\) or not.
5. In ∆ ABC, right-angled at B, AC + BC = 25 cm and AB = 5 cm. Determine the values of \(sin A\), \(cos A\) and \(tan A\).
6. In ∆ PQR, right-angled at Q (see Fig. 8.1), PQ = 3 cm and PR = 6 cm. Determine\(\angle QPR\) and \(\angle PRQ\).
7. Evaluate the following :
(i) \(sin 60^\circ cos 30^{\circ} + sin 30^\circ cos 60^\circ\)
(ii) \(2tan^2 45^\circ + cos^2 30^\circ – sin^2 60^\circ\)
8. If \(tan (A + B) =\sqrt 3\) and \(tan (A – B) = \frac{1}{\sqrt3}\); \(0^\circ < A + B ≤ 90^\circ\); A > B, find A and B.
9. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) \((cosec\theta – cot\theta)^2 =\frac {1 - cos}{1 + cos\theta}\)
(ii) \(\frac{cos A}{1 + sin A}+ \frac{1 + sin A}{cosA}= 2 sec A\)
(iii) \(\frac{tan\theta}{1 - cot\theta}+\frac{cot\theta}{1 - tan\theta} = 1 + sec\theta cosec\theta\)
(iv) \(\frac{1 + secnA}{secA} = \frac{sin^2A }{1 - cosA}\)
(v) \(\frac{cos A – sin A + 1}{cos A + sin A – 1}=cosec A + cot A\), using the identity \(cosec^2 A = 1 + cot^2 A\).
(vi) \(\sqrt{\frac{1 + sinA}{1 – sin A}}=sec A + tan A\)
(vii) \(\frac{sin\theta - 2sin^3\theta}{2cos^3\theta - cos\theta}=tan\theta\)
(viii) \((sin A + cosec A)^2 + (cos A + sec A)^2\) = 7 + \(tan^2 A + cot^2 A\)
(ix) \((cosec A - sin A)(sec A - cos A)=\frac{1}{tanA + cot A}\)