sinA+cosA=m & sinA-cosA=n
then prove that m^2+n^2=2
Answers
Answered by
0
m=cosA-sinA , m=cosA+sinA.
m
2
−n
2
m
2
+n
2
=
(cosA−sinA)
2
−(cosA+sinA)
2
(cosA−sinA)
2
(cosA+sinA)
2
=
(cosA−sinA+cosA+sinA)(cosA−sinA−cosA−sinA)
cos
2
A+sin
2
A−2cosAsinA+cos
2
A+sin
2
A+2cosAsinA
=
2cosA(−2sina)
1+1
=
−4cosAsinA
2
=
2
−1
secAcosA.
=
2cosAsinA
−1
=
2sin2A
−1
[∵sin2θ=2sinθcosθ]
=
1+tan
2
A
2tan4
−1
[∵sin2θ=
1+tan
2
θ
2tan
2
θ
]
=
2tanA
−(1+tan
2
A)
=
2
−1
(
tana
1
+
tanA
tan
2
A
)
=
2
−1
(cotA+tanA)
Answered by
2
sinA+cosA = m
m² =( sinA +cosA)²
= sin²A+2sinA.cosA + cos²A
= 1 + 2sinA.cosA.......................(i)
n² = (sinA-cosA)²
=sin²A-2sinAcosA +cos²A
=1 - 2sinA.cosA.........................(ii)
from equation (i) and (ii)
m²+ n² = 2 (proved)
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