Question

1+cos theta - sin theta /sin theta (1 +cos theta) =cot theta

Answer 1

Question:

Prove that:

$\displaystyle{\sf\:\dfrac{1\:+\:\cos\:\theta\:-\:\sin^2\:\theta}{\sin\:\theta\:(\:1\:+\:\cos\:\theta\:)}\:=\:\cot\:\theta}$

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{1\:+\:\cos\:\theta\:-\:\sin^2\:\theta}{\sin\:\theta\:(\:1\:+\:\cos\:\theta\:)} \:=\:\cot\:\theta}}}$

Step-by-step-explanation:

We have to prove a trigonometric equation.

$\displaystyle{\sf\:\dfrac{1\:+\:\cos\:\theta\:-\:\sin^2\:\theta}{\sin\:\theta\:(\:1\:+\:\cos\:\theta\:)}\:=\:\cot\:\theta}$

$\displaystyle{\sf\:LHS\:=\:\dfrac{1\:+\:\cos\:\theta\:-\:\sin^2\:\theta}{\sin\:\theta\:(\:1\:+\:\cos\:\theta\:)}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{1\:+\:\cos\:\theta\:-\:(\:1\:-\:\cos^2\:\theta\:)}{\sin\:\theta\:(\:1\:+\:\cos\:\theta\:)}\:\qquad\dots[\:\sin^2\:\theta\:+\:\cos^2\:\theta\:=\:1\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cancel{1}\:+\:\cos\:\theta\:-\:\cancel{1}\:+\:\cos^2\:\theta}{\sin\:\theta\:(\:1\:+\:\cos\:\theta\:)}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cos\:\theta\:+\:\cos^2\:\theta}{\sin\:\theta\:(\:1\:+\:\cos\:\theta\:)}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cos\:\theta}{\sin\:\theta\:(\:1\:+\:\cos\:\theta\:)}\:+\:\dfrac{\cos^2\:\theta}{\sin\:\theta\:(\:1\:+\:\cos\:\theta\:)}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cot\:\theta}{1\:+\:\cos\:\theta}\:+\:\dfrac{\cot\:\theta\:.\:\cos\:\theta}{1\:+\:\cos\:\theta}\:\qquad\dots\left[\:\cot\:\theta\:=\:\dfrac{\cos\:\theta}{\sin\:\theta}\:\right]}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cot\:\theta\:+\:\cot\:\theta\:\cos\:\theta}{1\:+\:\cos\:\theta}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cot\:\theta\:\cancel{(\:1\:+\:\cos\:\theta\:)}}{\cancel{(\:1\:+\:\cos\:\theta\:)}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\cot\:\theta}$

$\displaystyle{\implies\sf\:RHS\:=\:\cot\:\theta}$

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

Hence proved!