Question

Why (sin²x-sin²xcos²x)/cos²x =[sin²x(1-cos²x)]\cos²x

Answer 1

Answer:

Given,

f(x)=

⎩

⎪

⎨

⎪

⎧

x

2

+1

−1

cos

2

x−sin

2

x−1

,

k,

x



=0

x=0

If f(x) is continuous at x=0, then

x→0

lim

f(x)=f(0)

x→0

lim

x

2

+1

−1

cos

2

x−sin

2

x−1

=k

x→0

lim

x

2

+1

−1

(1−sin

2

x)−sin

2

x−1

=k

x→0

lim

x

2

+1

−1

−2sin

2

x

=k

x→0

lim

(

x

2

+1

−1)(

x

2

+1

+1)

−2sin

2

x(

x

2

+1

+1)

=k

x→0

lim

x

2

−2sin

2

x(

x

2

+1

+1)

=k

⇒−2

x→0

lim

x

2

sin

2

x(

x

2

+1

+1)

=k

⇒−2

x→0

lim

(

x

sinx

)

2

x→0

lim

(

x

2

+1

+1)=k

⇒k=−2×1×(1+1)=−4

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