Question

Prove that cosec theta/cosec theta -1+cosec theta/cosec theta+1=2/cos^2theta
Plz answer in LHS=RHS form

Answer 1

Question:

Prove that:

$\displaystyle\sf\:\dfrac{\csc\:\theta}{\csc\:\theta\:-\:1}\:+\:\dfrac{\csc\:\theta}{\csc\:\theta\:+\:1}\:=\:\dfrac{2}{\cos^2\:\theta}$

Answer:

$\displaystyle\boxed{\red{\sf\:\dfrac{\csc\:\theta}{\csc\:\theta\:-\:1}\:+\:\dfrac{\csc\:\theta}{\csc\:\theta\:+\:1}\:=\:\dfrac{2}{\cos^2\:\theta}}}$

Step-by-step-explanation:

We have to prove the given trigonometric equation.

Considering LHS of the given equation,

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\csc\:\theta}{\csc\:\theta\:-\:1}\:+\:\dfrac{\csc\:\theta}{\csc\:\theta\:+\:1}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\csc\:\theta\:\times\:(\:\csc\:\theta\:+\:1\:)\:+\:\csc\:\theta\:\times\:(\:\csc\:\theta\:-\:1\:)}{\:(\:\csc\:\theta\:-\:1\:)\:\times\:(\:\csc\:\theta\:+\:1\:)}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\csc^2\:\theta\:+\:1\:+\:\csc^2\:\theta\:-\:1}{\csc^2\:\theta\:-\:1^2}\:\:\:-\:-\:[\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:=\:a^2\:-\:b^2\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\csc^2\:\theta\:+\:\csc^2\:\theta}{\csc^2\:\theta\:-\:1}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2\:\csc^2\:\theta}{\cot^2\:\theta}\:\:\:-\:-\:[\:\because\:1\:+\:\cot^2\:\theta\:=\:\csc^2\:\theta\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2\:\times\:\left(\:\dfrac{1}{\sin\:\theta}\:\right)^2}{\cot^2\:\theta}\:\:\:-\:-\:-\:[\:\because\:\csc\:\theta\:=\:\dfrac{1}{\sin\:\theta}\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2\:\times\:\left(\:\dfrac{1}{\sin\:\theta}\:\right)^2}{\left(\:\dfrac{\cos\:\theta}{\sin\:\theta}\:\right)^2}\:\:\:-\:-\:-\:[\:\because\:\cot\:\theta\:=\:\dfrac{\cos\:\theta}{\sin\:\theta}\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2\:\times\:\left(\:\dfrac{1}{\sin\:\theta}\:\right)\:\times\:\left(\:\dfrac{1}{\sin\:\theta}\:\right)}{\left(\:\dfrac{\cos\:\theta}{\sin\:\theta}\:\right)\:\times\:\left(\:\dfrac{\cos\:\theta}{\sin\:\theta}\:\right)}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2\:\times\:\cancel{\sin\:\theta}\:\times\:\cancel{\sin\:\theta}}{\cancel{\sin\:\theta}\:\times\:\cancel{\sin\:\theta}\:\times\:\cos\:\theta\:\times\:\cos\:\theta}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2}{\cos^2\:\theta}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{2}{\cos^2\:\theta}}$

$\displaystyle{\therefore\boxed{\red{\sf\:LHS\:=\:RHS}}}$

Hence proved!

Answer 2

Answer:

$\bf{To\:Prove:}\sf\dfrac{\csc\:\theta}{\csc\:\theta\:-\:1}\:+\:\dfrac{\csc\:\theta}{\csc\:\theta\:+\:1} = \dfrac{2}{\cos^2\:\theta}$

⠀

$\bf{Proof:}$

$\implies\sf\:\dfrac{\csc\:\theta}{\csc\:\theta\:-\:1}\:+\:\dfrac{\csc\:\theta}{\csc\:\theta\:+\:1}\\\\\\\implies\sf\:\csc\:\theta\bigg(\dfrac{1}{\csc\:\theta\:-\:1}\:+\:\dfrac{1}{\csc\:\theta\:+\:1}\bigg)\\\\\\\implies\sf\:\csc\:\theta\bigg(\dfrac{\csc\:\theta\: + \:1 + \csc\:\theta\:-\:1}{(\csc\:\theta\:-\:1)(\csc\:\theta\: + \:1)}\bigg)\\\\\\\implies\sf \csc\:\theta \times \dfrac{2\csc\:\theta}{\csc^2\:\theta - 1}\\\\\\\implies\sf \dfrac{2\csc^2\:\theta}{\cot^2\:\theta}\\\\\\\implies\sf \dfrac{2 \times \frac{1}{\sin^2\:\theta}}{\frac{\cos^2\:\theta}{ \sin^2\:\theta} }\\\\\\\implies\sf \dfrac{2}{\cos^2\:\theta}$

$\rule{180}{1.5}$

$\boxed{\begin{minipage}{6 cm}\underline{\text{Trigonometric Identities Used Here :}}\\ \\{\textcircled{\footnotesize\textsf{1}}}\:\:(\csc^{2} \theta - 1)=\cot^2\theta\\ \\{\textcircled{\footnotesize\textsf{2}}} \: \:\csc^{2} \theta = \dfrac{1}{\sin^{2} \theta} \\\\ {\textcircled{\footnotesize\textsf{3}}} \: \:\cot^{2} \theta = \dfrac{\cos^{2} \theta}{\sin^{2} \theta} \end{minipage}}$