mathemerize

Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A.

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 »

If A, B and C are the interior angles of a triangle ABC, show that sin\(B+C\over 2\) = cos\(A\over 2\).

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 …

If A, B and C are the interior angles of a triangle ABC, show that sin\(B+C\over 2\) = cos\(A\over 2\). Read More »

Show that : (i) tan 48 tan 23 tan 42 tan 67 = 1 (ii) cos 38 cos 52 – sin 38 sin 52 = 0

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 »