Question

(cosec A - sin A)(sec A - cos A)= 1/
tan A + cot A​

Answer 1

Answer:

Hence Proved.

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Answer 2

Answer:

$\displaystyle{\boxed{\red{\sf\:(\:cosec\:A\:-\:\sin\:A\:)\:(\:\sec\:A\:-\:\cos\:A\:)\:=\:\dfrac{1}{\tan\:A\:+\:\cot\:A}}}}$

Step-by-step-explanation:

We have given a trigonometric equation.

We have to prove that equation.

$\displaystyle{\sf\:(\:cosec\:A\:-\:\sin\:A\:)\:(\:\sec\:A\:-\:\cos\:A\:)\:=\:\dfrac{1}{\tan\:A\:+\:\cot\:A}}$

$\displaystyle{\sf\:RHS\:=\:\dfrac{1}{\tan\:A\:+\:\cot\:A}}$

$\displaystyle{\implies\sf\:RHS\:=\:\dfrac{1}{\dfrac{\sin\:A}{\cos\:A}\:+\:\dfrac{\cos\:A}{\sin\:A}}\:\quad\:\:\:-\:-\:-\:\left[\:\because\:\tan\:A\:=\:\dfrac{\sin\:A}{\cos\:A}\:\&\:\cot\:A\:=\:\dfrac{\cos\:A}{\sin\:A}\:\right]}$

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

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

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

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

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

$\displaystyle{\implies\sf\:RHS\:=\:\sin\:A\:\cos\:A\:\times\:\left(\:\dfrac{\sin\:A\:\cos\:A}{\sin\:A\:\cos\:A}\:\right)\:\:-\:-\:-\:[\:Multiplying\:and\:dividing\:by\:\sin\:A\:\cos\:A\:]}$

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

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

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

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

$\displaystyle{\implies\sf\:RHS\:=\:(\:cosec\:A\:-\:\sin\:A\:)\:\left(\:\dfrac{1}{\cos\:A}\:-\:\cos\:A\:\right)\:\:-\:-\:-\:\left[\:\because\:cosec\:A\:=\:\dfrac{1}{\sin\:A}\:\right]}$

$\displaystyle{\implies\sf\:RHS\:=\:(\:cosec\:A\:-\:\sin\:A\:)\:(\:\sec\:A\:-\:\cos\:A\:)\:\:-\:-\:-\:\left[\:\because\:\sec\:A\:=\:\dfrac{1}{\cos\:A}\:\right]}$

$\displaystyle{\implies\sf\:LHS\:=\:(\:cosec\:A\:-\:\sin\:A\:)\:(\:\sec\:A\:-\:\cos\:A\:)}$

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

Hence proved!