2M If cos20= and sin20= 1-M2 then is 1+M 2 1+ M2 pla say to me
Answers
Answer:
Correct options are B) and C)
Let t=tan
2
2
θ
(m+2)sinθ+(2m−1)cosθ=2m+1
⇒(m+2)
1+t
2
2t
+(2m−1)
1+t
2
1−t
2
=2m+1
2mt+4t+2m−2mt
2
−1+t
2
=2m+2mt
2
+1+t
2
⇒(2t−1)(mt−1)=0
Gives t=
2
1
or t=
m
1
For t=tan
2
2
θ
=
2
1
⇒tanθ=
1−t
2
2t
=
1−
4
1
1
=
3
4
For t=tan
2
2
θ
=
m
1
⇒tanθ=
1−t
2
2t
=
1−
m
2
1
2
m
1
=2m(m
2
−1)
Answer:
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Question
If m
2
+m
′
2
+2mm
′
cosθ=1,
n
2
+n
′
2
+2nn
′
cosθ=1,
and mn+m
′
n
′
+(mn
′
+m
′
n)=cosθ=θ, then
prove that m
2
+n
2
=csc
2
θ.
Hard
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Solution
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The given relation can be written as
(m
′
+mcosθ
2
)+m
2
−m
2
cos
2
θ=1
or (m
′
+mcosθ
2
)=1−m
2
sin
2
θ
Similarly (n
′
+ncosθ)
2
=1−n
2
sin
2
θ
Now (m
′
+mcosθ)=(n
′
+ncosθ)
m
′
n
′
+(mn
′
+m
′
n)cosθ+mncos
2
θ
=−mn+mncos
2
θ by given relation
=−mn(1−cos
2
θ)=−mnsin
2
θ
Now squaring both sides, we get
or (m
′
+mcosθ)
2
(n
′
+ncosθ)
2
=m
2
n
2
sin
4
θ
Hence substituting from (1) and (2) in (3), we get
(1−m
2
sin
2
θ)(1−n
2
sin
2
θ)=m
2
n
2
sin
4
θ
or =(m
2
+n
2
)sin
2
θ=1 i.e. =m
2
+n
2
=csc
2
θ