it’s urgent→(1+sintheta-costheta/1+sintheta+costheta)²=1-costheta/1+costheta do it fast!!!!!!! About the author Kylie
[tex]\large\underline{\sf{Solution-}}[/tex] Consider, LHS [tex]\rm :\longmapsto\: {\bigg(\dfrac{1 + sin\theta – cos\theta }{1 + sin\theta + cos\theta } \bigg) }^{2} [/tex] can be rewritten as [tex] \rm \: = \: \: {\bigg(\dfrac{1 + sin\theta – cos\theta }{(1 + sin\theta) + cos\theta } \bigg) }^{2} [/tex] On rationalizing the denominator, we get [tex] \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} [/tex] We know, [tex] \red{ \boxed{ \sf{ (x + y)(x – y)\: = \: {x}^{2} – {y}^{2} }}}[/tex] [tex] \rm \: \: = \: {\bigg(\dfrac{ {(1 + sin\theta + cos\theta )}^{2} }{ {(1 + sin\theta )}^{2} – {cos}^{2}\theta } \bigg) }^{2} [/tex] We know, [tex] \red{ \boxed{ \sf{(x + y + z)^{2}\: = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx}}}[/tex] [tex] \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} [/tex] We know, [tex] \red{ \boxed{ \sf{ {sin}^{2}x + {cos}^{2} x \: = \: 1}}}[/tex] [tex] \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} [/tex] [tex] \rm \: \: = \: {\bigg(\dfrac{1 +1+ 2sin\theta- 2sin\theta cos\theta – 2cos\theta}{{sin}^{2}\theta + {sin}^{2}\theta + 2sin\theta } \bigg) }^{2} [/tex] [tex] \rm \: \: = \: {\bigg(\dfrac{2+ 2sin\theta- 2sin\theta cos\theta – 2cos\theta}{ 2{sin}^{2}\theta + 2sin\theta } \bigg) }^{2} [/tex] [tex] \rm \: \: = \: {\bigg(\dfrac{1+ sin\theta- sin\theta cos\theta – cos\theta}{ {sin}^{2}\theta + sin\theta } \bigg) }^{2} [/tex] [tex] \rm \: \: = \: {\bigg(\dfrac{(1+ sin\theta)- cos\theta( 1 + sin\theta)}{ {sin}\theta(1 + sin\theta) } \bigg) }^{2} [/tex] [tex] \rm \: \: = \: {\bigg(\dfrac{(1+ sin\theta)( 1 – cos\theta)}{ {sin}\theta(1 + sin\theta) } \bigg) }^{2} [/tex] [tex] \rm \: \: = \: {\bigg(\dfrac{ 1 – cos\theta}{ {sin}\theta } \bigg) }^{2} [/tex] [tex] \rm \: \: = \: \dfrac{ {(1 – cos\theta )}^{2} }{ {sin}^{2}\theta } [/tex] [tex] \rm \: \: = \: \dfrac{ {(1 – cos\theta )}^{2} }{ 1 – {cos}^{2}\theta } [/tex] [tex] \rm \: \: = \: \dfrac{ {(1 – cos\theta )}^{2} }{( 1 – {cos}\theta)(1 + cos\theta )} [/tex] [tex] \rm \: \: = \: \dfrac{1 – cos\theta }{1 + cos\theta } [/tex] 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 Reply
[tex]\large\underline{\sf{Solution-}}[/tex]
Consider, LHS
[tex]\rm :\longmapsto\: {\bigg(\dfrac{1 + sin\theta – cos\theta }{1 + sin\theta + cos\theta } \bigg) }^{2} [/tex]
can be rewritten as
[tex] \rm \: = \: \: {\bigg(\dfrac{1 + sin\theta – cos\theta }{(1 + sin\theta) + cos\theta } \bigg) }^{2} [/tex]
On rationalizing the denominator, we get
[tex] \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} [/tex]
We know,
[tex] \red{ \boxed{ \sf{ (x + y)(x – y)\: = \: {x}^{2} – {y}^{2} }}}[/tex]
[tex] \rm \: \: = \: {\bigg(\dfrac{ {(1 + sin\theta + cos\theta )}^{2} }{ {(1 + sin\theta )}^{2} – {cos}^{2}\theta } \bigg) }^{2} [/tex]
We know,
[tex] \red{ \boxed{ \sf{(x + y + z)^{2}\: = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx}}}[/tex]
[tex] \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} [/tex]
We know,
[tex] \red{ \boxed{ \sf{ {sin}^{2}x + {cos}^{2} x \: = \: 1}}}[/tex]
[tex] \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} [/tex]
[tex] \rm \: \: = \: {\bigg(\dfrac{1 +1+ 2sin\theta- 2sin\theta cos\theta – 2cos\theta}{{sin}^{2}\theta + {sin}^{2}\theta + 2sin\theta } \bigg) }^{2} [/tex]
[tex] \rm \: \: = \: {\bigg(\dfrac{2+ 2sin\theta- 2sin\theta cos\theta – 2cos\theta}{ 2{sin}^{2}\theta + 2sin\theta } \bigg) }^{2} [/tex]
[tex] \rm \: \: = \: {\bigg(\dfrac{1+ sin\theta- sin\theta cos\theta – cos\theta}{ {sin}^{2}\theta + sin\theta } \bigg) }^{2} [/tex]
[tex] \rm \: \: = \: {\bigg(\dfrac{(1+ sin\theta)- cos\theta( 1 + sin\theta)}{ {sin}\theta(1 + sin\theta) } \bigg) }^{2} [/tex]
[tex] \rm \: \: = \: {\bigg(\dfrac{(1+ sin\theta)( 1 – cos\theta)}{ {sin}\theta(1 + sin\theta) } \bigg) }^{2} [/tex]
[tex] \rm \: \: = \: {\bigg(\dfrac{ 1 – cos\theta}{ {sin}\theta } \bigg) }^{2} [/tex]
[tex] \rm \: \: = \: \dfrac{ {(1 – cos\theta )}^{2} }{ {sin}^{2}\theta } [/tex]
[tex] \rm \: \: = \: \dfrac{ {(1 – cos\theta )}^{2} }{ 1 – {cos}^{2}\theta } [/tex]
[tex] \rm \: \: = \: \dfrac{ {(1 – cos\theta )}^{2} }{( 1 – {cos}\theta)(1 + cos\theta )} [/tex]
[tex] \rm \: \: = \: \dfrac{1 – cos\theta }{1 + cos\theta } [/tex]
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