Solution : The value of tan 120 degrees is \(-\sqrt{3}\). Proof : We Know that tan 60 = \(\sqrt{3}\). By using formula of tan 2A, tan 120 = \(2 tan 60\over 1 – tan^2 60\) = \(2 \times \sqrt{3}\over 1 – 3\) \(\implies\)  tan 120 = \(-\sqrt{3}\)

Solution : The value of cos 120 degrees is \(-1\over 2\). Proof : We Know that sin 60 = \(\sqrt{3}\over 2\) and cos 60 = \(1\over 2\) By using cos 2A formula, cos 120 = \(cos^2 60\) – \(sin^2 60\) = \(1\over 4\) – \(3\over 4\) \(\implies\)  cos 120 = \(-1\over 2\)

Solution : The value of sin 120 degrees is \(\sqrt{3}\over 2\). Proof : We Know that sin 60 = \(\sqrt{3}\over 2\) and cos 60 = \(1\over 2\) By using the formula of sin 2A, sin 120 = 2 sin 60 cos 60 = 2 \(\times\) \(\sqrt{3}\over 2\) \(\times\) \(1\over 2\) \(\implies\)  sin 120 = \(\sqrt{3}\over

Solution : The formula of cos C – cos D is 2 sin(\(C + D\over 2\)) sin(\(D – C\over 2\)). Example : Express cos \(6\theta\) – cos \(8\theta\) as a product. Solution : We have, cos \(6\theta\) – cos \(8\theta\) = 2 sin(\(6\theta + 8\theta\over 2\)) sin(\(8\theta – 6 \theta\over 2\)) = 2 sin \(7\theta\) sin

Solution : The formula of cos C + cos D is 2 cos(\(C + D\over 2\)) cos(\(C – D\over 2\)). Example : Express cos \(4\theta\) + cos \(8\theta\) as a product. Solution : We have, cos \(4\theta\) + cos \(8\theta\) = 2 cos(\(8\theta + 4\theta\over 2\)) cos(\(8\theta – 4\theta\over 2\)) = 2 cos \(6\theta\) cos

Solution : The formula of sin C – sin D is 2 sin(\(C – D\over 2\)) cos(\(C + D\over 2\)). Example : Express sin \(6\theta\) – sin \(2\theta\) as a product. Solution : We have, sin \(6\theta\) – sin \(2\theta\) = 2 sin(\(6\theta – 2\theta\over 2\)) cos(\(6\theta + 2\theta\over 2\)) = 2 sin \(2\theta\) cos

Solution : The formula of sin C + sin D is 2 sin(\(C + D\over 2\)) cos(\(C – D\over 2\)). Example : Express sin \(4\theta\) + sin \(2\theta\) as a product. Solution : We have, sin \(4\theta\) + sin \(2\theta\) = 2 sin(\(4\theta + 2\theta\over 2\)) cos(\(4\theta – 2\theta\over 2\)) = 2 sin \(3\theta\) cos