====== Question 3, Exercise 10.3 ======
Solutions of Question 3 of Exercise 10.3 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 3(i)=====
Prove that $$\dfrac{\cos {{75}^{\circ }}+\cos {{15}^{\circ }}}{\sin {{75}^{\circ }}-\sin {{15}^{\circ }}}=\sqrt{3}.$$
====Solution====
We have identities:
$$\cos \alpha +\cos \beta =2\cos \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)$$
and
$$\sin \alpha -\sin \beta =2\cos \left( \dfrac{\alpha +\beta }{2} \right)\sin \left( \dfrac{\alpha -\beta }{2} \right).$$
Now
\begin{align}L.H.S.&=\dfrac{\cos {{75}^{\circ }}+\cos {{15}^{\circ }}}{\sin {{75}^{\circ }}-\sin {{15}^{\circ }}}\\
&=\dfrac{2\cos \left( \dfrac{{{75}^{\circ }}+{{15}^{\circ }}}{2} \right)\cos \left( \dfrac{{{75}^{\circ }}-{{15}^{\circ }}}{2} \right)}{2\cos \left( \dfrac{{{75}^{\circ }}+{{15}^{\circ }}}{2} \right)\sin \left( \dfrac{{{75}^{\circ }}-{{15}^{\circ }}}{2} \right)}\\ 
&=\dfrac{\cos 30^\circ}{\sin 30^\circ}\\
&=\dfrac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}=R.H.S.\end{align}

=====Question 3(ii)=====
Prove that $$\dfrac{\sin {{135}^{\circ }}-\cos {{120}^{\circ }}}{\sin {{135}^{\circ }}+\cos {{120}^{\circ }}}=3+2\sqrt{2}.$$
====Solution==== 
\begin{align}L.H.S.&=\dfrac{\sin {{135}^{\circ }}-\cos {{120}^{\circ }}}{\sin {{135}^{\circ }}+\cos {{120}^{\circ }}}\\ 
&=\dfrac{\dfrac{1}{\sqrt{2}}-\left( -\dfrac{1}{2} \right)}{\dfrac{1}{\sqrt{2}}+\left( -\dfrac{1}{2} \right)}\\ 
&=\dfrac{2+\sqrt{2}}{2-\sqrt{2}}\\
&=\dfrac{2+\sqrt{2}}{2-\sqrt{2}}\times \dfrac{2+\sqrt{2}}{2+\sqrt{2}}\\
&=\dfrac{6+4\sqrt{2}}{4-2}\\ 
&=3+2\sqrt{2}=R.H.S.\end{align}

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