Question

prove (b) no....☺☺☺☺

Answer 1

$\underline{\underline{\Huge{\textbf{Question:}}}}$

$\sf\textsf{Prove that:}$

$\LARGE{\mathbf{\dfrac{\tan\, \theta+\sin\, \theta}{\tan\, \theta-\sin\, \theta} = \dfrac{\sec\, \theta+1}{\sec\, \theta-1}}}$ ———[1]




$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\underline{\LARGE{\textsf{Step-1:}}}$

$\textsf{Consider LHS of Eq.[1]}$

$\underline{\Large{\textbf{LHS\: =\:}}}$

$\implies \mathbf{\dfrac{\tan\, \theta+\sin\, \theta}{\tan\, \theta-\sin\, \theta}}$




$\underline{\LARGE{\textsf{Step-2:}}}$

$\textsf{Express tan in terms of sin and cos}$

$\quad\LARGE{\boxed{\quad\bigstar\;\; \mathbf{\tan\, \theta = \dfrac{\sin\, \theta}{\cos\, \theta} \quad}}}$

$\implies \mathbf{\dfrac{\left(\dfrac{\sin \: \theta}{\cos\, \theta} \right) +\sin\, \theta}{\left(\dfrac{\sin\, \theta}{\cos\, \theta} \right)-\sin\, \theta}}$




$\underline{\LARGE{\textsf{Step-3:}}}$

$\textsf{Combine the terms in both Numerator}\\\sf\textsf{and Denominator and simplify}$

$\implies \mathsf{\dfrac{ \dfrac{\sin\, \theta + \sin\, \theta \cdot \cos \, \theta}{\cos \: \theta}}{\dfrac{\sin\, \theta - \sin\, \theta \cdot \cos \theta}{\cos \: \theta}}}$

$\implies \mathsf{\dfrac{\sin\, \theta + \sin\, \theta \cdot \cos \, \theta}{\cancel{\cos \: \theta}}\times \dfrac{\cancel{\cos\, \theta}}{\sin\, \theta - \sin\, \theta \cdot \cos \, \theta}}$

$\implies \mathsf{\dfrac{\sin\, \theta+\sin\, \theta\cdot\cos\, \theta}{\sin\, \theta-\sin\, \theta\cdot\cos\, \theta}}$




$\underline{\LARGE{\textsf{Step-4:}}}$

$\textsf{Take \sin\, \theta common in both Numerator}\\\sf\textsf{and Denominator to cancel}$

$\implies \mathsf{\dfrac{\cancel{\sin\, \theta}\, (1+\cos\, \theta)}{\cancel{\sin\, \theta}\, (1-\cos\, \theta)}}$

$\implies \mathsf{\dfrac{1+\cos\, \theta}{1-\cos\, \theta}}$




$\underline{\LARGE{\textsf{Step-5:}}}$

$\sf\textsf{Express cos in terms of sec}$

$\quad\LARGE{\boxed{\quad\bigstar\;\; \mathbf{\cos\, \theta = \dfrac{1}{\sec\, \theta} \quad}}}$

$\implies \mathsf{\dfrac{1+ \left(\dfrac{1}{\sec\, \theta}\right)}{1-\left(\dfrac{1}{\sec\, \theta}\right)}}$




$\underline{\LARGE{\textsf{Step-6:}}}$

$\textsf{Combine the terms in both Numerator}\\\sf\textsf{and Denominator and simplify}$

$\implies \dfrac{\dfrac{\sec\, \theta+1}{\sec\, \theta}}{ \dfrac{\sec\, \theta-1}{\sec \: \theta}}$

$\implies \mathsf{\dfrac{\sec\, \theta+1}{\sec\, \theta-1}\times \dfrac{\cancel{\sec\, \theta}}{\cancel{\sec\, \theta}}}$

$\implies \mathsf{\dfrac{\sec\, \theta+1}{\sec\, \theta-1}}$

$\underline{\Large{\textbf{\: =\:RHS}}}$






$\blacksquare\; \; \underline{\LARGE{\textbf{Hence Proved}}}$