Question 8, Exercise 10.1

Solutions of Question 8 of Exercise 10.1 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: $\tan \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$

\begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}+\theta \right)\\ &=\dfrac{\sin \left( \dfrac{\pi }{4}+\theta \right)}{\cos \left( \dfrac{\pi }{4}+\theta \right)}\\ &=\dfrac{\sin\dfrac{\pi }{4}\cos \theta +\cos\dfrac{\pi }{4}\sin\theta}{\cos\dfrac{\pi}{4}\cos \theta -\sin \dfrac{\pi }{4}\sin \theta }\\ &=\dfrac{\dfrac{1}{\sqrt{2}}\cos \theta +\dfrac{1}{\sqrt{2}}\sin\theta}{\dfrac{1}{\sqrt{2}}\cos \theta -\dfrac{1}{\sqrt{2}}\sin \theta }\\ &=\dfrac{\dfrac{1}{\sqrt{2}}\left(\cos\theta +\sin\theta\right)}{\dfrac{1}{\sqrt{2}}\left(\cos \theta -\sin\theta\right)}\\ &=\dfrac{\cos \theta +\sin\theta }{\cos\theta -\sin\theta }=R.H.S.\end{align}

Prove that: $\tan \left( \dfrac{\pi }{4}-\theta \right)=\dfrac{1-tan\theta }{1+tan\theta }$

\begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}-\theta \right)\\ &=\dfrac{\sin \left( \dfrac{\pi }{4}-\theta \right)}{\cos \left( \dfrac{\pi }{4}-\theta \right)}\\ &=\dfrac{\sin \dfrac{\pi }{4}\cos \theta -\sin \theta \cos \dfrac{\pi }{4}}{\cos \dfrac{\pi }{4}\cos \theta +\sin \dfrac{\pi }{4}\sin \theta }\\ &=\dfrac{\dfrac{1}{\sqrt{2}}\cos \theta -\sin \theta \dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}\cos \theta +\dfrac{1}{\sqrt{2}}\sin \theta }\\ &=\dfrac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }\\ &=\dfrac{\cos \theta \left( 1-\dfrac{\sin \theta }{\cos \theta } \right)}{\cos \theta \left( 1+\dfrac{\sin \theta }{\cos \theta } \right)}\\ &=\dfrac{\left( 1-\tan \theta \right)}{\left( 1+\tan \theta \right)}\\ &=R.H.S.\end{align}

Alternative Method

\begin{align}L.H.S.&=\tan\left( \dfrac{\pi }{4}-\theta \right)\\ &=\dfrac{\tan\frac{\pi }{4}-\tan\theta}{1+\tan\frac{\pi }{4}\tan\theta}\\ &=\dfrac{1-\tan\theta}{1+1\cdot\tan\theta } \quad \because \tan\dfrac{\pi}{4}=1\\ &=\dfrac{\left( 1-\tan \theta \right)}{\left( 1+\tan \theta \right)}\\ &=R.H.S.\end{align}

Prove that: $\dfrac{\tan \left( \alpha +\beta \right)}{\cot \left( \alpha -\beta \right)}=\dfrac{{{\tan }^{2}}\alpha -{{\tan }^{2}}\beta }{1-{{\tan }^{2}}\alpha {{\tan }^{2}}\beta }$

\begin{align}\tan (\alpha +\beta )&=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ \cot (\alpha -\beta )&=\dfrac{1+\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }\\ L.H.S.&=\dfrac{\tan \left( \alpha +\beta \right)}{\cot \left( \alpha -\beta \right)}\\ &=\dfrac{\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }}{\dfrac{1+\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }}\\ &=\dfrac{\left( \tan \alpha +\tan \beta \right)\left( \tan \alpha -\tan \beta \right)}{\left( 1+\tan \alpha \tan \beta \right)\left( 1-\tan \alpha \tan \beta \right)}\\ &=\dfrac{{{\tan }^{2}}\alpha -{{\tan }^{2}}\beta }{1-{{\tan }^{2}}\alpha {{\tan }^{2}}\beta }\\ &=R.H.S.\end{align}

Prove that: $\dfrac{1-\tan \theta \tan \phi }{1+\tan \theta \tan \phi }=\dfrac{\cos \left( \theta +\phi \right)}{\cos \left( \theta -\phi \right)}$

\begin{align}L.H.S.&=\dfrac{1-\tan \theta \tan \phi }{1+\tan \theta \tan \phi }\\ &=\dfrac{1-\dfrac{\sin \theta }{\cos \theta }\dfrac{\sin \phi }{\cos \phi }}{1+\dfrac{\sin \theta }{\cos \theta }\dfrac{\sin \phi }{\cos \phi }}\\ &=\dfrac{\dfrac{\cos \theta \cos \phi -\sin \theta \sin \phi }{\cos \theta \cos \phi }}{\dfrac{\cos \theta \cos \phi +\sin \theta \sin \phi }{\cos \theta \cos \phi }}\\ &=\dfrac{\cos \theta \cos \phi -\sin \theta \sin \phi }{\cos \theta \cos \phi +\sin \theta \sin \phi }\\ &=\dfrac{\cos \left( \theta +\phi \right)}{\cos \left( \theta -\phi \right)}\\ &=R.H.S.\end{align}