Question

under root 1+cosA by 1-cosA =1+cosA by sinA

​

Answer 1

TO PROVE :–

$\\ \implies\rm \sqrt\dfrac{1 + \cos(A) }{1 - \cos(A) } = \dfrac{1 + \cos(A) }{ \sin(A) }$

SOLUTION :–

• Let's take L.H.S. –

$\\ \rm \: \: = \sqrt\dfrac{1 + \cos(A) }{1 - \cos(A) } \: \:$

• Now Rationalization of denominator –

$\\ \rm \: \: = {\sqrt{\dfrac{1 + \cos(A) }{1 - \cos(A) } \times\dfrac{1 + \cos(A) }{1 + \cos(A)}} }\: \:$

$\\ \rm \: \: = {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{ \{1 - \cos(A) \} \{ 1 + \cos(A)\}}}}$

• We know that –

$\\ \large\longrightarrow{ \boxed{ \rm(a - b)(a + b) = {a}^{2} - {b}^{2} }}$

• So that –

$\\ \rm \: \: = {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{ \{1 - \cos^{2} (A) \}}}}$

• We also know that –

$\\ \large\bf\implies \sin^{2}( \theta) +\cos^{2}( \theta) = 1$

$\\ \large\bf\implies \sin^{2}( \theta)= 1 - \cos^{2}( \theta)$

$\\ \large\bf\implies 1 - \cos^{2}( \theta) = \sin^{2}( \theta)$

• So that –

$\\ \rm \: \: = \:\: {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{\sin^{2} (A)}}}$

$\\ \rm \: \: = \: \: \dfrac{ 1 + \cos(A)}{\sin(A)}$

$\\ \rm \: \: = \: \: R.H.S.$

$\\ \large \longrightarrow{ \boxed{ \rm Hence \: \:Proved }}$

Answer 2

Step-by-step explanation:

TO PROVE :–

\begin{gathered} \\ \implies\rm \sqrt\dfrac{1 + \cos(A) }{1 - \cos(A) } = \dfrac{1 + \cos(A) }{ \sin(A) } \\ \end{gathered}

⟹

1−cos(A)

1+cos(A)

=

sin(A)

1+cos(A)

SOLUTION :–

• Let's take L.H.S. –

\begin{gathered} \\ \rm \: \: = \sqrt\dfrac{1 + \cos(A) }{1 - \cos(A) } \: \: \\ \end{gathered}

=

1−cos(A)

1+cos(A)

• Now Rationalization of denominator –

\begin{gathered} \\ \rm \: \: = {\sqrt{\dfrac{1 + \cos(A) }{1 - \cos(A) } \times\dfrac{1 + \cos(A) }{1 + \cos(A)}} }\: \: \\ \end{gathered}

=

1−cos(A)

1+cos(A)

×

1+cos(A)

1+cos(A)

\begin{gathered} \\ \rm \: \: = {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{ \{1 - \cos(A) \} \{ 1 + \cos(A)\}}}}\\ \end{gathered}

=

{1−cos(A)}{1+cos(A)}

{1+cos(A)}

2

• We know that –

\begin{gathered} \\ \large\longrightarrow{ \boxed{ \rm(a - b)(a + b) = {a}^{2} - {b}^{2} }}\\ \end{gathered}

⟶

(a−b)(a+b)=a

2

−b

2

• So that –

\begin{gathered} \\ \rm \: \: = {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{ \{1 - \cos^{2} (A) \}}}}\\ \end{gathered}

=

{1−cos

2

(A)}

{1+cos(A)}

2

• We also know that –

\begin{gathered} \\ \large\bf\implies \sin^{2}( \theta) +\cos^{2}( \theta) = 1\\ \end{gathered}

⟹sin

2

(θ)+cos

2

(θ)=1

\begin{gathered} \\ \large\bf\implies \sin^{2}( \theta)= 1 - \cos^{2}( \theta) \\ \end{gathered}

⟹sin

2

(θ)=1−cos

2

(θ)

\begin{gathered} \\ \large\bf\implies 1 - \cos^{2}( \theta) = \sin^{2}( \theta)\\ \end{gathered}

⟹1−cos

2

(θ)=sin

2

(θ)

• So that –

\begin{gathered} \\ \rm \: \: = \:\: {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{\sin^{2} (A)}}}\\ \end{gathered}

=

sin

2

(A)

{1+cos(A)}

2

\begin{gathered} \\ \rm \: \: = \: \: \dfrac{ 1 + \cos(A)}{\sin(A)}\\ \end{gathered}

=

sin(A)

1+cos(A)

\begin{gathered} \\ \rm \: \: = \: \: R.H.S. \\ \end{gathered}

=R.H.S.

\begin{gathered} \\ \large \longrightarrow{ \boxed{ \rm Hence \: \:Proved }}\\ \end{gathered}

⟶

HenceProved