Question

$\textbf{Prove the identity :}\ \textless \ br /\ \textgreater \ [tex] {\Large({\frac{1 \: + \: sin \: \theta \: - \: cos \: \theta}{1 \: + \: sin \: \theta \: + \: cos \: \theta})}^{2} \: = \: \frac{1 \: - \: cos \: \theta}{1 \: + \: cos \: \theta} }$
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Answer 1

Question :

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

Solution :

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

Let us assume

1 as a

sin∅ as b

cos ∅ as c

So for the numerator we can use the identity (a + b – c)²

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

And for denominator we will apply (a + b + c)²

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

$\sf \frac{(a + b – c)² }{(a + b + c)²} = \frac{1 - c}{ 1 + c}$

$\sf \frac{a² + b² + c² + 2(ab – bc – ca)}{a² + b² + c² + 2(ab + bc + ca)} = \frac{1 - c}{ 1 + c}$

$\sf \frac{1² + sin²\theta + cos²\theta + 2(1×sin\theta - sin\theta × cos\theta - cos\theta×1)}{1² + sin²\theta + cos²\theta + 2(1×sin\theta + sin\theta × cos\theta + cos\theta×1)} = \frac{1 - cos\theta}{ 1 + cos\theta}$

Now we know that sin²∅ + cos²∅ = 1

so by applying that we get

$\sf \frac{1 + 1+ 2(1×sin\theta - sin\theta × cos\theta - cos\theta×1)}{1+1 + 2(1×sin\theta + sin\theta × cos\theta + cos\theta×1)} = \frac{1 - cos\theta}{ 1 + cos\theta}$

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

$\sf \frac{\cancel{4}(\cancel{sin\theta }- \cancel{sin\theta cos\theta} - cos\theta)}{\cancel{4}(\cancel{sin\theta }+\cancel{sin\theta cos\theta} + cos\theta)} = \frac{1 - cos\theta}{ 1 + cos\theta}$

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

Hence Proved !!

Answer 2

Answer:

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