# Question 2 and 3, Review Exercise 10

Solutions of Question 2 and 3 of Review Exercise 10 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Prove the identity $\dfrac{2\sin \theta \sin 2\theta }{\cos \theta +\cos 3\theta }=\tan 2\theta \tan \theta$.

\begin{align}L.H.S.&=\dfrac{2\sin \theta \sin 2\theta }{\cos \theta +\cos 3\theta }\\ &=\dfrac{2\sin \theta \sin 2\theta }{\cos 3\theta +\cos \theta }\\ &=\dfrac{2\sin \theta \sin 2\theta }{2\cos \dfrac{3\theta +\theta }{2}\cos \dfrac{3\theta -\theta }{2}}\\ &=\dfrac{2\sin \theta \sin 2\theta }{2\cos 2\theta \cos \theta }\\ &=\tan \theta \tan 2\theta =R.H.S.\end{align}

Prove the identity $\dfrac{\sin 10a-\sin 4a}{\sin 4a+\sin 2a}=\dfrac{\cos 7a}{\cos a}$.

\begin{align}L.H.S.&=\dfrac{\sin 10a-\sin 4a}{\sin 4a+\sin 2a}\\ &=\dfrac{2\cos \left( \dfrac{10a+4a}{2} \right)\sin \left( \dfrac{10a-4a}{2} \right)}{2\sin \dfrac{4a+2a}{2}\cos \dfrac{4a-2a}{2}}\\ &=\dfrac{\cos 7a\sin 3a}{\sin 3a\cos a}\\ &=\dfrac{\cos 7a}{\cos a}=R.H.S.\end{align}