Trigonometry Questions

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

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 »

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 »