# 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.

Prove that $$\dfrac{\cos {{75}^{\circ }}+\cos {{15}^{\circ }}}{\sin {{75}^{\circ }}-\sin {{15}^{\circ }}}=\sqrt{3}.$$

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}

Prove that $$\dfrac{\sin {{135}^{\circ }}-\cos {{120}^{\circ }}}{\sin {{135}^{\circ }}+\cos {{120}^{\circ }}}=3+2\sqrt{2}.$$

\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}