Author name: mathemerize

Solution : The formula of 2 sin A sin B is cos (A – B) – cos (A + B). Example : Solve : 2 sin 75 sin 15. Solution : We have, 2 sin 75 sin 15 = cos (75 – 15) – cos (75 + 15) = cos 60 – cos 90 = […]

Solution : The formula of 2 cos A cos B is cos (A + B) + cos (A – B). Example : Solve : 2 cos 75 cos 15. Solution : We have, 2 cos 75 cos 15 = cos (75 + 15) + cos (75 – 15) = cos 90 + cos 60 =

Solution : The formula of 2 cos A sin B is sin (A + B) – sin (A – B). Example : Solve : 2 cos 75 sin 15. Solution : We have, 2 cos 75 sin 15 = sin (75 + 15) – sin (75 – 15) = sin 90 – sin 60 =

Solution : The formula of 2 sin A cos B is sin (A + B) + sin (A – B). Example : Solve : 2 sin 75 cos 15. Solution : We have, 2 sin 75 cos 15 = sin (75 + 15) + sin (75 – 15) = sin 90 + sin 60 =

Solution : The value of tan 75 degrees is \(\sqrt{3} + 1\over \sqrt{3} – 1\). Proof :  We write tan 75 as tan (45 + 30). Using the formula of tan (A + B), tan 75 = tan (45 + 30) = \(tan 45 + tan 30\over 1 – tan 45 tan 30\) \(\implies\) tan

Solution : The value of tan 15 degrees is \(\sqrt{3} – 1\over \sqrt{3} + 1\). Proof :  We write tan 15 as tan (45 – 30). Using the formula of tan (A – B), tan 15 = tan (45 – 30) = \(tan 45 – tan 30\over 1 + tan 45 tan 30\) \(\implies\) tan

Solution : The formula of tan (A + B + C) is \(tan A + tan B + tan C – tan A tan B tan C\over 1 – tan A tan B – tan B tan C – tan C tan A\). Proof : We have, tan (A + B + C) = tan((A

Solution : The formula of cos(A + B + C) is cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C. Proof :  We have, cos (A + B + C) = cos ((A + B) + C)

Solution : The formula of sin (A + B + C) is sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C. Proof :  We have, sin (A + B + C) = sin ((A + B) +

Solution : We have, cos (A + B) cos (A – B) = (cos A cos B – sin A sin B) (cos A  cos B + sin A sin B) = \(cos^2 A cos^2 B\) – \(sin^2 A sin^2 B\) = \(cos^2 A (1 – sin^2 B)\) – \((1 – cos^2 A) sin^2 B\)