Question

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(1+sintheta-costheta/1+sintheta+costheta)²=1-costheta/1+costheta

do it fast!!!!!!!
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Answer 1

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\: {\bigg(\dfrac{1 + sin\theta - cos\theta }{1 + sin\theta + cos\theta } \bigg) }^{2}$

can be rewritten as

$\rm \: = \: \: {\bigg(\dfrac{1 + sin\theta - cos\theta }{(1 + sin\theta) + cos\theta } \bigg) }^{2}$

On rationalizing the denominator, we get

$\rm \: = \: \: {\bigg(\dfrac{1 + sin\theta - cos\theta }{(1 + sin\theta) + cos\theta } \times \dfrac{1 + sin\theta - cos\theta }{(1 + sin\theta) - cos\theta } \bigg) }^{2}$

We know,

$\red{ \boxed{ \sf{ (x + y)(x - y)\: = \: {x}^{2} - {y}^{2} }}}$

$\rm \: \: = \: {\bigg(\dfrac{ {(1 + sin\theta + cos\theta )}^{2} }{ {(1 + sin\theta )}^{2} - {cos}^{2}\theta } \bigg) }^{2}$

We know,

$\red{ \boxed{ \sf{(x + y + z)^{2}\: = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx}}}$

$\rm \: \: = \: {\bigg(\dfrac{1 + {sin}^{2}\theta + {cos}^{2}\theta + 2sin\theta - 2sin\theta cos\theta - 2cos\theta }{1 + {sin}^{2}\theta + 2sin\theta - {cos}^{2}\theta } \bigg) }^{2}$

We know,

$\red{ \boxed{ \sf{ {sin}^{2}x + {cos}^{2} x \: = \: 1}}}$

$\rm \: \: = \: {\bigg(\dfrac{1 +({sin}^{2}\theta + {cos}^{2}\theta)+ 2sin\theta- 2sin\theta cos\theta - 2cos\theta}{(1 - {cos}^{2}\theta) + {sin}^{2}\theta + 2sin\theta } \bigg) }^{2}$

$\rm \: \: = \: {\bigg(\dfrac{1 +1+ 2sin\theta- 2sin\theta cos\theta - 2cos\theta}{{sin}^{2}\theta + {sin}^{2}\theta + 2sin\theta } \bigg) }^{2}$

$\rm \: \: = \: {\bigg(\dfrac{2+ 2sin\theta- 2sin\theta cos\theta - 2cos\theta}{ 2{sin}^{2}\theta + 2sin\theta } \bigg) }^{2}$

$\rm \: \: = \: {\bigg(\dfrac{1+ sin\theta- sin\theta cos\theta - cos\theta}{ {sin}^{2}\theta + sin\theta } \bigg) }^{2}$

$\rm \: \: = \: {\bigg(\dfrac{(1+ sin\theta)- cos\theta( 1 + sin\theta)}{ {sin}\theta(1 + sin\theta) } \bigg) }^{2}$

$\rm \: \: = \: {\bigg(\dfrac{(1+ sin\theta)( 1 - cos\theta)}{ {sin}\theta(1 + sin\theta) } \bigg) }^{2}$

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

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

$\rm \: \: = \: \dfrac{ {(1 - cos\theta )}^{2} }{ 1 - {cos}^{2}\theta }$

$\rm \: \: = \: \dfrac{ {(1 - cos\theta )}^{2} }{( 1 - {cos}\theta)(1 + cos\theta )}$

$\rm \: \: = \: \dfrac{1 - cos\theta }{1 + cos\theta }$

Hence, Proved

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1