====== Question 7, Exercise 10.1 ======

Solutions of Question 1 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.
=====Question 7(i)=====
Show that:   $\cot \left( \alpha +\beta  \right)=\dfrac{\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }$ 
====Solution====
\begin{align}L.H.S.&=\cot (\alpha +\beta )\\
&=\dfrac{1}{\tan (\alpha +\beta )}\\
&=\dfrac{1}{\,\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\,}\\
&=\dfrac{1-\tan \alpha \tan \beta }{\,\tan \alpha +\tan \beta \,}\\
&=\dfrac{\tan \alpha \tan \beta \left( \dfrac{1}{\tan \alpha \tan \beta }-1 \right)}{\,\tan \alpha \tan \beta \left( \dfrac{1}{\tan \beta }+\dfrac{1}{\tan \alpha } \right)\,}\\
&=\dfrac{\cot \alpha \,\,\cot \beta -1}{\,\cot \beta +\cot \alpha \,}\\
&=\dfrac{\,\cot \alpha \,\,\cot \beta -1\,}{\,\cot \alpha +\cot \beta \,}\\
&= R.H.S.\end{align}

=====Question 7(ii)===== 
Show that:   $\dfrac{\sin \left( \alpha +\beta  \right)}{\cos \alpha \cos \beta }=\tan \alpha +\tan \beta .$ 
====Solution====
\begin{align}L.H.S.&=\dfrac{\sin \left( \alpha +\beta  \right)}{\cos \alpha \cos \beta }\\ 
&=\dfrac{\sin \alpha \cos \beta +\sin \beta \cos \alpha }{\cos \alpha \cos \beta }\\ 
&=\dfrac{\sin \alpha \cos \beta }{\cos \alpha \cos \beta }+\dfrac{\sin \beta \cos \alpha }{\cos \alpha \cos \beta }\\
&=\dfrac{\sin \alpha }{\cos \alpha }+\dfrac{\sin \beta }{\cos \beta }\\
&=\tan \alpha +\tan \beta \\ 
&=R.H.S.\end{align} 

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