Trigonometry Questions

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 – …

Prove the following identities, where the angles involved are acute angles for which the expressions are defined : Read More »

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 …

Evaluate : Read More »

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 …

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 »

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