If cosec θ - sin θ=a and sec θ - cos θ=b , then find the value of (a2b)23+(ab2)23.
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Asked on November 22, 2019 by
Satvik Coelho
If cosecθ−sinθ=a
3
and secθ−cosθ=b
3
, prove that a
2
b
2
(a
2
+b
2
)=1.
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ANSWER
Given : cosecθ−sinθ=a
3
and secθ−cosθ=b
3
⇒
sinθ
1
−sinθ=a
3
⇒
sinθ
cos
2
θ
=a
3
⇒a=(
sinθ
cos
2
θ
)
3
1
Similarly ; b=(
cosθ
sin
2
θ
)
3
1
⇒a
2
b
2
(a
2
+b
2
)=a
4
b
2
+a
2
b
4
⇒(
sinθ
cos
2
θ
)
3
4
(
cosθ
sin
2
θ
)
3
2
+(
sinθ
cos
2
θ
)
3
2
(
cosθ
sin
2
θ
)
3
4
⇒(cosθ)
8/3−2/3
.(sinθ)
4/3−4/3
+(sinθ)
8/3−2/3
.(cosθ)
4/3−4/3
⇒cos
2
θ+sin
2
θ=1
⇒a
2
b
2
(a
2
+b
2
)=1
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