Solution : (i) L.H.S = tan 48 tan 23 tan 42 tan 67 = \(1\over cot 48\). tan 23 tan 42 \(1\over cot 67\) = \(1\over cot (90 – 42)\). tan 23 tan 42 \(1\over cot (90 – 23)\) = \(1\over tan 42\). tan 23 tan 42 \(1\over tan 23\) = 1 = R.H.S. (ii) …
Show that : (i) tan 48 tan 23 tan 42 tan 67 = 1 (ii) cos 38 cos 52 – sin 38 sin 52 = 0 Read More »
Question : (i) \(sin 18\over cos 72\) (ii) \(tan 26\over cot 64\) (iii) cos 48 – sin 42 (iv) cosec 31 – sec 59 Solution : (i) \(sin 18\over cos 72\) = \(sin 18\over cos (90 – 18)\) = \(sin 18\over sin 18\) = 1 (ii) \(tan 26\over cot 64\) = \(tan 26\over cot (90 …
Question : (i) Sin (A + B) = sin A + sin B (ii) The value of \(sin \theta\) increases as \(\theta\) increases. (iii) The value of \(cos \theta\) increases as \(\theta\) increases. (iv) \(sin \theta\) = \(cos \theta\) for all values of \(\theta\). (v) Cot A is not defined for A = 0. Solution …
State whether the following are true or false. Justify you answer. Read More »
Solution : We have, tan (A + B) = \(\sqrt{3}\) \(\implies\) tan (A + B) = tan 60 \(\implies\) A + B = 60 …………(1) Also, tan (A – B) = \(1\over \sqrt{3}\) \(\implies\) tan(A – B) = tan 30 \(\implies\) A – B = 30 …………(2) …
Question : (i) \(2 tan 30\over 1 + tan^2 30\) = (a) sin 60 (b) cos 60 (c) tan 60 (d) sin 30 (ii) \(1 – tan^2 45\over 1 + tan^2 45\) = (a) tan 90 (b) 1 (c) sin 45 (d) 0 (iii) sin 2A = 2 sin A is the true when A …
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Question : (i) sin 60 cos 30 + cos 60 sin 30 (ii) \(2 tan^2 45\) + \(cos^2 30\) – \(sin^2 60\) (iii) \(cos 45\over sec 30 + cosec 30\) (iv) \(sin 30 + tan 45 – cosec 60\over sec 30 + cos 60 + cot 45\) (v) \(5 cos^2 60 + 4 sec^2 30 …
Question : State whether the following are true or false. Justify the answer. (i) The value of tan A is always less than 1. (ii) sec A = \(12\over 5\) for some values of angle A. (iii) cos A is the abbreviation used for the cosecant of angle A. (iv) cot A is the product …
State whether the following are true or false. Justify the answer. Read More »
Solution : We have, PQ = 5 cm PR + QR = 25 cm ………..(1) In triangle PQR, By Pythagoras Theorem, \({PR}^2\) = \({PQ}^2\) + \({QR}^2\) \(\implies\) \({PQ}^2\) = \({PR}^2\) – \({QR}^2\) \(\implies\) \({PQ}^2\) = (PR + QR)(PR – QR) \(\implies\) \(5^2\) = (PR – QR). 25 …
Solution : Consider a \(\triangle\) ABC, in which \(\angle\) B = 90 For \(\angle\) A, we have : Base = AB, Perp. = BC, and Hyp. = AC, tan A = \(\perp\over base\) = \(BC\over AB\) = \(1\over \sqrt{3}\) Let BC = k and AB = \(\sqrt{3} k, AC = \(\sqrt{{AB}^2 + {BC}^2}\) = 2k …
Solution : We have, 3 cot A = 4 \(\implies\) cot A = \(4\over 3\) = \(AB\over BC\) Let AB = 4k, then BC = 3k By Pythagoras Theorem, \({AC}^2\) = \({AB}^2\) + \({BC}^2\) \(\implies\) \({AC}^2\) = \(25k^2\) \(\implies\) AC = 5k Thus, tan A = \(BC\over AB\) = \(3k\over 4k\) = \(3\over 4\) …
