Question

prove that sinA/1+cosA = 1-cosA/sinA​

Answer 1

Answer:

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

Step-by-step-explanation:

We have given a trigonometric equation.

We have to prove that equation.

The given trigonometric equation is

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

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

Multiplying and dividing by ( 1 + cos A ), we get,

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

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

We know that,

$\displaystyle{\boxed{\blue{\sf\:(\:a\:-\:b\:)\:(\:a\:+\:b\:)\:=\:a^2\:-\:b^2\:}}}$

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

We know that,

$\displaystyle{\boxed{\pink{\sf\:\sin^2\:A\:+\:\cos^2\:A\:=\:1\:}}}$

$\displaystyle{\therefore\sf\:1\:-\:\cos^2\:A\:=\:\sin^2\:A}$

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

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

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

$\displaystyle{\sf\:LHS\:=\:\dfrac{\sin\:A}{1\:+\:\cos\:A}}$

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

Hence proved!

Answer 2

Step-by-step explanation:

Answer:

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

1+cosA

sinA

=

sinA

1−cosA

Step-by-step-explanation:

We have given a trigonometric equation.

We have to prove that equation.

The given trigonometric equation is

\displaystyle{\sf\:\dfrac{\sin\:A}{1\:+\:\cos\:A}\:=\:\dfrac{1\:-\:\cos\:A}{\sin\:A}}

1+cosA

sinA

=

sinA

1−cosA

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

sinA

1−cosA

Multiplying and dividing by ( 1 + cos A ), we get,

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

sinA

1−cosA

×

1+cosA

1+cosA

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

sinA(1+cosA)

(1−cosA)(1+cosA)

We know that,

(a−b)(a+b)=a

2

−b

2

1−cos

2

A

We know that

2

A+cos

2

A=1

2

A=sin

2

A

1+cosA

sinA

LHS=RHS

Hence proved!