प्र04 संधि विच्छेद कीजिए-
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Answers
Answer:
Appropriate Question
\begin{gathered}\sf \: If \: tanA + sinA = m \: and \: tanA-sinA=n \\ \sf \: show \: that \: {m}^{2} \: - \: {n}^{2} \: = \: 4 \: \sqrt{mn} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}
IftanA+sinA=mandtanA−sinA=n
showthatm
2
−n
2
=4
mn
\large\underline{\sf{Solution-}}
Solution−
Given that,
\rm :\longmapsto\:m = tanA + sinA:⟼m=tanA+sinA
and
\rm :\longmapsto\:n = tanA - sinA:⟼n=tanA−sinA
Now, Consider
\rm :\longmapsto\: {m}^{2} - {n}^{2}:⟼m
2
−n
2
On substituting the values of m and n, we get
\rm \: = \: \: {(tanA + sinA)}^{2} - {(tanA - sinA)}^{2} = (tanA+sinA)
2
−(tanA−sinA)
2
We know
\boxed{ \bf{ \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}
(x+y)
2
−(x−y)
2
=4xy
So, using this identity, we get
\rm \: = \: \: 4 \: tanA \: sinA = 4tanAsinA
\bf\implies \: {m}^{2} - {n}^{2} = 4 \: tanA \: sinA - - - (1)⟹m
2
−n
2
=4tanAsinA−−−(1)
Now, Consider
\rm :\longmapsto\:4 \sqrt{mn}:⟼4
mn
On substituting the values of m and n, we get
\rm \: = \: \: 4 \sqrt{(tanA + sinA)(tanA - sinA)} = 4
(tanA+sinA)(tanA−sinA)
We know,
\boxed{ \bf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}
(x+y)(x−y)=x
2
−y
2
So, using this identity we get,
\rm \: = \: \: 4 \sqrt{ {tan}^{2}A - {sin}^{2}A } = 4
tan
2
A−sin
2
A
\rm \: = \: \: 4 \sqrt{ \dfrac{ {sin}^{2} A}{ {cos}^{2} A} - {sin}^{2}A } = 4
cos
2
A
sin
2
A
−sin
2
A
\rm \: = \: \: 4 \sqrt{ \bigg(\dfrac{1}{ {cos}^{2} A} - 1\bigg){sin}^{2}A } = 4
(
cos
2
A
1
−1)sin
2
A
We know,
\boxed{ \bf{ \: \frac{1}{cosx} = sinx}}
cosx
1
=sinx
So, using this
\rm \: = \: \: 4 \sqrt{ \bigg( {sec}^{2}A - 1\bigg){sin}^{2}A } = 4
(sec
2
A−1)sin
2
A
We know,
\boxed{ \bf{ \: {sec}^{2}x - {tan}^{2}x = 1}}
sec
2
x−tan
2
x=1
So, using this, we get
\rm \: = \: \: 4 \sqrt{ {tan}^{2} A \times {sin}^{2} A} = 4
tan
2
A×sin
2
A
\rm \: = \: \: 4tanA \: sinA = 4tanAsinA
\bf\implies \:4 \sqrt{mn} = 4 \: tanA \: sinA - - - (2)⟹4
mn
=4tanAsinA−−−(2)
From equation (1) and (2), we concluded that
\boxed{ \bf{ \: {m}^{2} - {n}^{2} = 4 \sqrt{mn}}}
m
2
−n
2
=4
mn
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
Explanation:
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