Question

(1+sinθ+cosθ/1+sinθ−cosθ​)²=1+cosθ/1−cosθ​

Answer 1

Step-by-step explanation:

$\rm LHS = [ \frac{1 + sin \theta - cos \theta}{1 + sin \theta + cos \theta} ]^{2}$

$\rm = \big[ \frac{(1 + sin \theta) - cos \theta}{(1 + sin \theta )+ cos \theta} \big]^{2}$

On rationalise the denominator :-

$\rm = \big[ \frac{(1 + sin \theta) - cos \theta}{(1 + sin \theta )+ cos \theta} \times \frac{{(1 + sin \theta) - cos \theta}}{{(1 + sin \theta) - cos \theta}} \big]^{2}$

$\rm = \big[ \frac{((1 + sin \theta) - cos \theta)({(1 + sin \theta) - cos \theta)}}{((1 + sin \theta )+ cos \theta){((1 + sin \theta) - cos \theta)}}\big]^{2}$

$\rm = \big[ \frac{(1 + sin \theta)((1 + sin \theta) - cos \theta){ - cos \theta((1 + sin \theta) - cos \theta)}}{(1 + sin \theta )^{2} - (cos \theta)^{2} {}}\big]^{2}$

$\rm = \big[ \frac{(1 + sin \theta)(1 + sin \theta) - cos(1 + sin \theta){ - cos \theta(1 + sin \theta) - cos \theta( - cos \theta)}}{1 + sin^{2} \theta + 2sin \theta - cos^{2} \theta {}}\big]^{2}$

$\rm = \big[ \frac{1 + sin^{2} \theta+ 2sin \theta- cos \theta - cos \theta.sin \theta{ - cos \theta - cos \theta.sin \theta + cos^{2} \theta}}{1- cos^{2} \theta + sin^{2} \theta + 2sin \theta }\big]^{2}$

$\rm = \big[ \frac{1 + (sin^{2} \theta+ cos^{2} \theta)+ 2sin \theta- cos \theta - cos \theta- cos \theta.sin \theta- cos \theta.sin \theta }{ sin^{2} \theta + sin^{2} \theta + 2sin \theta }\big]^{2}$

$\rm = \big[ \frac{1 +1+ 2sin \theta-2 cos \theta - 2cos \theta.sin \theta }{ 2sin^{2} \theta + 2sin \theta }\big]^{2}$

$\rm = \big[ \frac{2+ 2sin \theta - 2cos \theta.sin \theta -2 cos \theta }{ 2sin^{2} \theta + 2sin \theta }\big]^{2}$

$\rm = \big[ \frac{2(1 + sin \theta )- 2cos \theta(sin \theta + 1) }{ 2sin \theta (sin \theta + 1) }\big]^{2}$

$\rm = \big[ \frac{(2- 2cos \theta)(1 + sin \theta ) }{ 2sin \theta (sin \theta + 1) }\big]^{2}$

$\rm = \big[ \frac{ \cancel2(1- cos \theta)(1 + sin \theta ) }{ \cancel2sin \theta (sin \theta + 1) }\big]^{2}$

$\rm = \big[ \frac{(1- cos \theta) \cancel{(1 + sin \theta ) }}{ sin \theta \cancel{ (sin \theta + 1)} }\big]^{2}$

$\rm = \big[ \frac{(1- cos \theta)}{ sin \theta }\big]^{2}$

$\rm = \frac{(1- cos \theta)^{2} }{ sin^{2} \theta }$

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

$\rm = \frac{ \cancel{(1- cos \theta)}(1- cos \theta) }{ \cancel{(1 - cos \theta)}(1 + cos \theta) } \: \: \{ \because \: {a}^{2} - {b}^{2} = (a - b)(a + b) \}$

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

$\rm = RHS$

Hence proved