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}