Question

$\textbf{Prove the identity :}$
${\Large({\frac{1 \: + \: sin \: \theta \: - \: cos \: \theta}{1 \: + \: sin \: \theta \: + \: cos \: \theta})}^{2} \: = \: \frac{1 \: - \: cos \: \theta}{1 \: + \: cos \: \theta} }$​

Answer 1

LHS(After taking square)

then

$\frac{2(1+ sin\theta -cos\theta -sin\theta.cos\theta)}{2(1+ sin\theta +cos\theta +sin\theta.cos\theta}$(Since $sin^{2}\theta + cos^{2} \theta = 1$)

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

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

$\frac{1-cos\theta}{1+ cos\theta}$(Prooved)

Answer 2

Answer:

cos theta ne chika isiliye sin theta ne usko pita..lol

Step-by-step explanation:

LHS(After taking square)

then

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

2(1+sinθ+cosθ+sinθ.cosθ

2(1+sinθ−cosθ−sinθ.cosθ)

(Since sin^{2}\theta + cos^{2} \theta = 1sin

2

θ+cos

2

θ=1 )

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

(1+sinθ)+cosθ(1+sinθ)

(1+sinθ)−cosθ(1+sinθ)

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

(1+sinθ)(1+cosθ)

(1+sinθ)(1−cosθ)

\frac{1-cos\theta}{1+ cos\theta}

1+cosθ

1−cosθ

(Prooved)

CODING NHI HO PAYA THICK SEE (T_T)..XD