Question

prove that 1+cosA +sinA/1+cosA-sinA= 1+sinA/cosA
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Answer 1

Answer:

Answer: The proof is done below.

Step-by-step explanation: We are given to prove the following trigonometric equality :

\dfrac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}=\dfrac{1+\sin\theta}{\cos\theta}.

sinθ−1+cosθ

sinθ+1−cosθ

=

cosθ

1+sinθ

.

We will be using the following formulas :

\begin{gathered}(i)~\dfrac{\sin\theta}{\cos\theta}=\tan\theta,\\\\\\(ii)~\sec\theta=\dfrac{1}{\cos\theta},\\\\\\(iii)~\sec^2\theta-\tan^2\theta=1\\\\(iii)~a^2-b^2=(a+b)(a-b).\end{gathered}

(i)

cosθ

sinθ

=tanθ,

(ii) secθ=

cosθ

1

,

(iii) sec

2

θ−tan

2

θ=1

(iii) a

2

−b

2

=(a+b)(a−b).

The proof is as follows :

\begin{gathered}\dfrac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}\\\\\\=\dfrac{\dfrac{\sin\theta+1-\cos\theta}{\cos\theta}}{\dfrac{\sin\theta-1+\cos\theta}{\cos\theta}}\\\\\\=\dfrac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}\\\\\\=\dfrac{\tan\theta+\sec\theta-(\sec^2\theta-\tan^2\theta)}{\tan\theta-\sec\theta+1}\\\\\\=\dfrac{\tan\theta+\sec\theta-(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{\tan\theta-\sec\theta+1}\\\\\\=\dfrac{(\tan\theta+\sec\theta)(1-\sec\theta+\tan\theta)}{1-\sec\theta+\tan\theta}\\\\\\=\tan\theta+\sec\theta\\\\\\=\dfrac{sin\theta}{\cos\theta}+\dfrac{1}{\cos\theta}\\\\\\=\dfrac{1+\sin\theta}{\cos\theta}\\\\=R.H.S.\end{gathered}

sinθ−1+cosθ

sinθ+1−cosθ

=

cosθ

sinθ−1+cosθ

cosθ

sinθ+1−cosθ

=

tanθ−secθ+1

tanθ+secθ−1

=

tanθ−secθ+1

tanθ+secθ−(sec

2

θ−tan

2

θ)

=

tanθ−secθ+1

tanθ+secθ−(secθ+tanθ)(secθ−tanθ)

=

1−secθ+tanθ

(tanθ+secθ)(1−secθ+tanθ)

=tanθ+secθ

=

cosθ

sinθ

+

cosθ

1

=

cosθ

1+sinθ

=R.H.S.

Thus, we get

\dfrac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}=\dfrac{1+\sin\theta}{\cos\theta}.

sinθ−1+cosθ

sinθ+1−cosθ

=

cosθ

1+sinθ

.

Hence proved.

⁹

Answer 2

Answer:

$refer \: to \: the \: attachment$