Please solve that's both questions about the mathamatics
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Answered by
5
Answer:
Equations 1 and 2 form pair of linear
equations in two variables r and p. Solving equation 1 and 2 we get
r=-3b
p=-4b
78]
2
Now,
r/p=-3b/-4b-3/4
Answered by
1
Answer:
Give(cosA−sinA)(secA−CosA)=
tanA+cotA
1
L.H.S.(coscA sinA)(secA-cosA)
⇒L.H.S=(
sinA
1
−sinA)(
cosA
1
−cosA)
⇒L.H.S=(
sinA
1−sin
2
A
)(
cosA
1−cos
2
A
)
⇒L.H.S=
sina.cosA
cos
2
A.sin
2
A
[∵sin
2
A+cos
2
A−1cos
2
A−1−sin
2
Asin
2
A=1−cos
2
A]
⇒L.H.S=
1
sinA.cosA
⇒L.H.S=
sin
2
A+cos
2
A
sinA.cosA
⇒L.H.S=
sinA.cosA
sin
2
A+cos
2
A
1
⇒L.H.S=
sinAcosA
sin
2
A
+
sinA.cosA
cos
2
A
1
⇒L.H.S=
cosA
sinA
+
sinA
cosA
1
⇒L.H.S=
tanA+cotA
1
=R.H.S
⇒L.H.S = R.H.S
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