Question

cosA/1-tanA+sin²A/sinA-cosA=sinA+cosA​

Answer 1

Question:

Prove that:

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

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{\cos\:A}{1\:-\:\tan\:A}\:+\:\dfrac{\sin^2\:A}{\sin\:A\:-\:\cos\:A}\:=\:\sin\:A\:+\:\cos\:A\:}}}$

Step-by-step-explanation:

We have given a trigonometric equation.

We have to prove that equation.

The given trigonometric equation is

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

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

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cos\:A}{1\:-\:\dfrac{\sin\:A}{\cos\:A}}\:+\:\dfrac{\sin^2\:A}{\sin\:A\:-\:\cos\:A}\:\qquad\cdots\left[\:\because\:\tan\:A\:=\:\dfrac{\sin\:A}{\cos\:A}\:\right]}$

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

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cos\:A\:\times\:\cos\:A}{\cos\:A\:-\:\sin\:A}\:+\:\dfrac{\sin^2\:A}{\sin\:A\:-\:\cos\:A}}$

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

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

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

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

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

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

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

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

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

Hence proved!

Answer 2

Answer:

Step-by-step explanation:

Hope it helps you