Question

prove that (1+sin theta - cos theta /1 + sin theta + cos theta )²= 1-cos theta/1+cos theta​

Answer 1

TO PROVE :-

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

SOLUTION :-

$\\ \sf \: L.H.S = { \left( \dfrac{1 + sin\theta - cos\theta}{1 + sin\theta + cos\theta}\right)}^{2} \\ \\ \\ \implies \sf \: \dfrac{ {(1 + sin\theta - cos\theta)}^{2} }{ {(1 + sin\theta + cos\theta)}^{2} }$

For numerator ,

★ (a-b)² = a² + 2ab + b²

• a = 1 + sinø
• b = cosø

For denominator ,

★ (a+b)² = a² + 2ab + b²

• a = 1 + sinø
• b = cosø

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

★ cos²ø = 1 - sin²ø

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

★ a² - b² = (a+b)(a-b)

• a = 1
• b = sinø

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

Taking (1+sinø) common ,

$\\ \implies \sf \: \dfrac{ \cancel{(1 + sin\theta)}(1 + \cancel{sin\theta} - 2cos\theta + 1 - \cancel{sin\theta})}{ \cancel{(1 + sin\theta)}(1 + \cancel{sin\theta }+ 2cos\theta + 1 - \cancel{sin\theta})} \\ \\ \\ \implies \sf \: \dfrac{2 - 2cos\theta}{ 2+ 2cos\theta}$

Taking 2 common ,

$\\ \implies \sf \: \dfrac{ \cancel2(1 - cos\theta)}{ \cancel2(1 + cos\theta)} \\ \\ \\ \implies \sf \: \dfrac{1 - cos\theta}{1 + cos\theta} = R.H.S \: \: \: \: \: \: (verified)$