Given cot 0 =7/8
then evaluate (i)
(1+sine)(1 - sino)
(1+cos)(1 - cose)
(1+sin o)
cos
8
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1
Answer:
Given,
cotθ=
8
7
tanθ=
cotθ
1
=
7
8
We know that,
tanθ=
adjacentSide
oppositeSide
From Pythagoras theorem,
Hypotenuse
2
=OppositeSide
2
+AdjacentSide
2
Hypotenuse
2
=8
2
+7
2
Hypotenuse
2
=64+49=113
Hypotenuse=
113
sinθ=
Hypotenuse
oppositeSide
=
113
8
cosθ=
Hypotenuse
AdjacentSide
=
113
7
Solution(i):
(1+cosθ)(1−cosθ)
(1+sinθ)(1−sinθ)
We have, a
2
−b
2
=(a+b)(a−b)
Similarly,
(1−sin
2
θ)=(1+sinθ)(1−sinθ)
(1−cos
2
θ)=(1+cosθ)(1−cosθ)
Therefore,
(1+cosθ)(1−cosθ)
(1+sinθ)(1−sinθ)
=
(1−cos
2
θ)
(1−sin
2
θ)
=
(1−(
113
7
)
2
)
(1−(
113
8
)
2
)
=
(113−49)
(113−64)
=
64
49
Solution(ii):
Given,
cotθ=
8
7
cot
2
θ=(
8
7
)
2
=
64
49
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