====== 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.
=====Question 4=====
Prove the identity ${{\sin }^{2}}\dfrac{\theta }{2}=\dfrac{\sin \theta \tan \dfrac{\theta }{2}}{2}$.
====Solution==== 
\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} 

=====Question 5=====
Prove the identity $\tan \theta \tan \dfrac{\theta }{2}=\sec \theta -1$.
====Solution====
\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} 
====Go to====
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