# Question 4 & 5, Review Exercise 10

Solutions of Question 4 & 5 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 ${{\sin }^{2}}\dfrac{\theta }{2}=\dfrac{\sin \theta \tan \dfrac{\theta }{2}}{2}$.

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

Prove the identity $\tan \theta \tan \dfrac{\theta }{2}=\sec \theta -1$.

\begin{align}L.H.S.&=\tan \theta \tan \dfrac{\theta }{2}\\ &\dfrac{\sin \theta }{\cos \theta }\dfrac{\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}}\\ &\dfrac{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}{\cos \theta }\dfrac{\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}}\quad (\because \sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2} )\\ &=\dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{\cos \theta }\\ R.H.S.&=\sec \theta -1\\ &=\dfrac{1}{\cos \theta }-1\\ &=\dfrac{1-\cos \theta }{\cos \theta }\\ &=\dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{\cos \theta }\quad (\because {{\sin }^{2}}\dfrac{\theta }{2}=\dfrac{1-\cos \theta }{2} )\\ L.H.S.&=R.H.S\end{align}