Question

1 + sin theta minus cos theta by 1 + sin theta + cos theta whole square equal to 1 minus cos theta by 1 + cos theta​

Answer 1

Step-by-step explanation:

LHS

(\frac{1+sin\theta -cos\theta}{1+sin\theta + cos\theta })^2(

1+sinθ+cosθ

1+sinθ−cosθ

)

2

Expending Numerator and Denominator using

(a + b + c )² = a² + b² + c² + 2ab + 2bc +2ca

\begin{lgathered}\frac{1+sin^2 \theta + cos^2 \theta + 2 sin\theta -2sin\theta cos\theta - 2 cos\theta}{1+sin^2\theta + cos^2\theta + 2 sin\theta +2sin\theta cos\theta +2 cos\theta} \\\end{lgathered}

1+sin

2

θ+cos

2

θ+2sinθ+2sinθcosθ+2cosθ

1+sin

2

θ+cos

2

θ+2sinθ−2sinθcosθ−2cosθ

∵ sin² Ф + cos² Ф = 1

\begin{lgathered}\frac{1+ 1 + 2 sin\theta -2sin\theta cos\theta - 2 cos\theta}{1+1 + 2 sin\theta +2sin\theta cos\theta +2 cos\theta} \\\end{lgathered}

1+1+2sinθ+2sinθcosθ+2cosθ

1+1+2sinθ−2sinθcosθ−2cosθ

\begin{lgathered}\frac{2 + 2 sin\theta -2sin\theta cos\theta - 2 cos\theta}{2 + 2 sin\theta +2sin\theta cos\theta +2 cos\theta} \\\end{lgathered}

2+2sinθ+2sinθcosθ+2cosθ

2+2sinθ−2sinθcosθ−2cosθ

Taking 2 common from first two terms and 2 cosФ from last two terms in both numerator and denominator

\begin{lgathered}\frac{2(1 + sin\theta) -2cos\theta(1+sin\theta)}{2(1 + sin\theta) +2cos\theta(1+sin\theta)} \\\end{lgathered}

2(1+sinθ)+2cosθ(1+sinθ)

2(1+sinθ)−2cosθ(1+sinθ)

\frac{(1+sin\theta)(2 - 2 cos\theta)}{(1+sin\theta)(2+2cos\theta)}

(1+sinθ)(2+2cosθ)

(1+sinθ)(2−2cosθ)

\begin{lgathered}\frac{2(1-cos\theta)}{2(1+cos\theta)} \\\end{lgathered}

2(1+cosθ)

2(1−cosθ)

\frac{(1-cos\theta)}{(1+cos\theta)} = RHS

(1+cosθ)

(1−cosθ)

=RHS