Question

Prove that:
sqr root of (1 + cos A
1-cos A)=tan A+sin A/tan A.sin A​

Answer 1

Question:

Prove that:

$\displaystyle{\sf\:\sqrt{\dfrac{1\:+\:\cos\:A}{1\:-\:\cos\:A}}\:=\:\dfrac{\tan\:A\:+\:\sin\:A}{\tan\:A\:.\:\sin\:A}}$

Answer:

$\displaystyle{\boxed{\red{\sf\:\sqrt{\dfrac{1\:+\:\cos\:A}{1\:-\:\cos\:A}}\:=\:\dfrac{\tan\:A\:+\:\sin\:A}{\tan\:A\:.\:\sin\:A}}}}$

Step-by-step-explanation:

We have given a trigonometric equation.

We have to prove that equation.

Now,

$\displaystyle{\sf\:\sqrt{\dfrac{1\:+\:\cos\:A}{1\:-\:\cos\:A}}\:=\:\dfrac{\tan\:A\:+\:\sin\:A}{\tan\:A\:.\:\sin\:A}}$

$\displaystyle{\sf\:RHS\:=\:\dfrac{\tan\:A\:+\:\sin\:A}{\tan\:A\:.\:\sin\:A}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\left(\:\dfrac{\sin\:A}{\cos\:A}\:\right)\:+\:\sin\:A}{\tan\:A\:.\:\sin\:A}\:\:\:-\:-\:-\:\left[\:\because\:\tan\:A\:=\:\dfrac{\sin\:A}{\cos\:A}\:\right]}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\dfrac{\sin\:A\:+\:\sin\:A\:.\:\cos\:A}{\cos\:A}}{\tan\:A\:.\:\sin\:A}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\dfrac{\sin\:A\:+\:\sin\:A\:.\cos\:A}{\cos\:A}}{\dfrac{\sin\:A}{\cos\:A}\:\times\:\sin\:A}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\dfrac{\sin\:A\:+\:\sin\:A\:.\cos\:A}{\cos\:A}}{\dfrac{\sin^2\:A}{\cos\:A}}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\sin\:A\:+\:\sin\:A\:.\:\cos\:A}{\cancel{\cos\:A}}\:\times\:\dfrac{\cancel{\cos\:A}}{\sin^2\:A}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\sin\:A\:+\:\sin\:A\:.\:\cos\:A}{\sin^2\:A}}$

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

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\sin\:A\:(\:1\:+\:\cos\:A\:)}{(\:1\:-\:\cos^2\:A\:)}\:\:\:-\:-\:-\:[\:\because\:\sin^2\:A\:+\:\cos^2\:A\:=\:1\:]}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\sin\:A\:(\:1\:+\:\cos\:A\:)}{1^2\:-\:(\:\cos\:A\:)^2}}$

$\displaystyle{\implies}\sf\:RHS\:=\:\dfrac{\sin\:A\:\cancel{(\:1\:+\:\cos\:A\:)}}{\cancel{(\:1\:+\:\cos\:A\:)}\:(\:1\:-\:\cos\:A\:)}\:\:\:\:-\:-\:-\:[\:\because\:a^2\:-\:b^2\:=\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:]$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\sin\:A}{1\:-\:\cos\:A}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\sqrt{1\:-\:\cos^2\:A}}{1\:-\:\cos\:A}\:\:\:-\:-\:-\:[\:\because\:\sin^2\:A\:+\:\cos^2\:A\:=\:1\:]}$

$\displaystyle{\implies\sf\:RHS\:=\:\left(\:\dfrac{\sqrt{1\:-\:\cos^2\:A}}{1\:-\:\cos\:A}\:\right)^2\:\:\:-\:-\:-\:[\:Taking\:square\:]}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{\sqrt{(\:1\:-\:\cos^2\:A\:)}^2}{(\:1\:-\:\cos\:A\:)^2}\:\:\:-\:-\:-\:\left[\:\because\:\left(\:\dfrac{a}{b}\:\right)^m\:=\:\dfrac{a^m}{b^m}\:\right]}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{1\:-\:\cos^2\:A}{(\:1\:-\:\cos\:A\:)\:\times\:(\:1\:-\:\cos\:A\:)}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{(\:1\:+\:\cos\:A\:)\:\times\:\cancel{(\:1\:-\:\cos\:A\:)}}{(\:1\:-\:\cos\:A\:)\:\times\:\cancel{(\:1\:-\:\cos\:A\:)}}\:\:\:\:-\:-\:-\:[\:\because\:a^2\:-\:b^2\:=\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:]}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{1\:+\:\cos\:A}{1\:-\:\cos\:A}}$

$\displaystyle{\implies\sf\:RHS\:=\:\sqrt{\dfrac{1\:+\:\cos\:A}{1\:-\:\cos\:A}}\:\:\:-\:-\:[\:Taking\:square\:roots\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:\sqrt{\dfrac{1\:+\:\cos\:A}{1\:-\:\cos\:A}}}$

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

Hence proved!