If y=(sin 8 +cosec0)2 + (cos 0 + sec )?, then minimum
value of y is :
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ANSWER
Let y=asecx+bcosec x
Therefore,
dx
dy
=asecxtanx−bcosec xcotx
When
dx
dy
=0,
Then asecxtanx=bcosec xcotx
cos
2
x
asinx
=
sin
2
x
bcosx
asin
3
x=bcos
3
x
tan
3
x=
a
b
tanx=(
a
b
)
1/3
x=tan
−1
(
a
b
)
1/3
⟹secx=
a
1/3
a
2/3
+b
2/3
,cosecx=
b
1/3
a
2/3
+b
2/3
Minimum value
y=a×secx+bcosec x
y=
a
1/3
a×
a
2/3
+b
2/3
+
b
1/3
b×
a
2/3
+b
2/3
y=(a
2/3
+b
2/3
)
3/2
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