Question

If cosec A = root 2 ,find the value of 2sin2A + 3cot2 A / 4(tan2 A - cos 2 A).

Answer 1

Answer:

2

Step-by-step explanation:

$cosecA = \sqrt{2}$. Therefore, $sinA=\frac{1}{\sqrt{2}}$

$A=45^0$

$cosA=cos45^0=\frac{1}{\sqrt{2}}$

$tanA=tan45^0=1$

$cotA=cot45^0=1$

Therefore, $2sin^2A + 3cot^2A=2(\frac{1}{\sqrt{2} })^2+3(1)^2=4$

$4(tan^2 A-cos^2A)=4(1^2-(\frac{1}{\sqrt{2}})^2)=4(1-\frac{1}{2})=2$

Hence, $\frac{2sin^2A + 3cot^2 A}{4(tan^2 A - cos^2A)}=\frac{4}{2}=2$

Answer 2

Answer:

2

Step-by-step explanation:

A=45

0

cosA=cos45^0=\frac{1}{\sqrt{2}}cosA=cos45

0

=

2

1

tanA=tan45^0=1tanA=tan45

0

=1

cotA=cot45^0=1cotA=cot45

0

=1

Therefore, 2sin^2A + 3cot^2A=2(\frac{1}{\sqrt{2} })^2+3(1)^2=42sin

2

A+3cot

2

A=2(

2

1

)

2

+3(1)

2

=4

4(tan^2 A-cos^2A)=4(1^2-(\frac{1}{\sqrt{2}})^2)=4(1-\frac{1}{2})=24(tan

2

A−cos

2

A)=4(1

2

−(

2

1

)

2

)=4(1−

2

1

)=2

Hence, \frac{2sin^2A + 3cot^2 A}{4(tan^2 A - cos^2A)}=\frac{4}{2}=2

4(tan

2

A−cos

2

A)

2sin

2

A+3cot

2

A

=

2

4

=2