Question

If y=(sin 8 +cosec0)2 + (cos 0 + sec )?, then minimum
value of y is :

Answer 1

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