Question : (i) \(1 – cos\theta\over 1 + cos\theta\) = \(({cosec\theta – cot\theta})^2\) (ii) \(1 + sin\theta\over cos\theta\) + \(cos\theta\over 1 + sin\theta\) = \(2sec\theta\) (iii) \(tan\theta\over 1 – cot\theta\) + \(cot\theta\over 1 – tan\theta\) = \(sec\theta + cosec\theta\) + 1 (iv) \(1 + sec\theta\over sec\theta\) = \(sin^2\theta\over 1 – cos\theta\) (v) \(cos A – …
Question : Choose the correct option. Justify your choice : (i) \(9 sec^2 A – 9 tan^2 A\) = (a) 1 (b) 9 (c) 8 (d) 0 (ii) \((1 – tan \theta + sec \theta)\)\((1 + cot \theta – cosec \theta)\) = (a) 0 (b) 1 (c) 2 (d) -1 (iii) (sec A + tan …
Choose the correct option. Justify your choice : Read More »
Question : Evaluate : (i) \(sin^2 63 + sin^2 27\over cos^2 17 + cos^2 73\) (ii) sin 25 cos 65 + cos 25 sin 65 Solution : (i) \(sin^2 63 + sin^2 27\over cos^2 17 + cos^2 73\) = \(sin^2 (90 – 27) + sin^2 27\over cos^2 17 + cos^2 (90 – 17)\) = \(cos^2 …
Solution : Consider a triangle ABC, in which \(\angle\) B = 90 For \(\angle\) A, we have : Base = AB, Perp = BC and Hyp = AC \(\therefore\) sec A = \(Hyp\over Base\) = \(AC\over AB\) So, \(AC\over AB\) = sec A = \(sec A\over 1\) Let AB = k and AC = …
Write the other trigonometric ratios of A in terms of sec A. Read More »
Solution : We know that \(cosec^2 A\) = 1 + \(cot^2 A\) \(\implies\) \(1\over sin^2 A\) = 1 + \(cot^2 A\) \(\implies\) \(sin^2 A\) = \(1\over 1 + cot^2 A\) \(\implies\) sin A = \(1\over \sqrt{1 + cot^2 A}\) Also, we know that \(sec^2 A\) = 1 + \(tan^2 A\) \(\implies\) \(sec^2 A\) = 1 …
Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A. Read More »
Solution : We have, sin 67 + cos 75 = sin(90 – 23) + cos(90 – 15) = cos 23 + sin 15
Solution : Since A, B and C are the interior angles of a triangle ABC \(\therefore\) A + B + C = 180 \(\implies\) \(A\over 2\) + \(B\over 2\) + \(C\over 2\) = 90 \(\implies\) \(B\over 2\) + \(C\over 2\) = 90 – \(A\over 2\) \(\implies\) sin(\(B+C\over 2\)) = sin(90 – \(A\over 2\)) \(\implies\) sin\(B+C\over …
Solution : Given, sec 4A = cosec(A – 20) \(\implies\) cosec(90 – 4A) = cosec(A – 20) \(\implies\) 90 – 4A = A – 20 \(\implies\) 5A = 110 \(\implies\) A = 22
Solution : We have, tan A = cot B \(\implies\) tan A = tan(90 – B) \(\implies\) A = 90 – B \(\implies\) A + B = 90
Solution : Given, tan 2A = cot(A – 18) \(\implies\) cot(90 – 2A) = cot(A – 18) \(\implies\) 90 – 2A = A – 18 \(\implies\) 3A = 108 \(\implies\) A = 36
