Question

ans plz...............​

Answer 1

Note:

I have used A instead of thetha.

Please do check it and apologize for the inconvenience caused.

Thank you... ^_^

Given:

tan A = √3 and 2 + cos ^ 2 A / 2 - sin ^ 2 A

To find:

The value of 2 + cos ^ 2 A / 2 - sin ^ 2 A

Solution:

tan A = √3

When A is 60° then, tan 60° = √3.

Therefore, A = 60°

Now, Substitute A = 60° in 2 + cos ^ 2 A / 2 - sin ^ 2 A

2 + cos ^ 2 A / 2 - sin ^ 2 A

2 + cos ^ 2 60° / 2 - sin ^ 2 60°

2 + (1 / 2)² / 2 - (√3 / 2)²

(2 + 1 / 4) / (2 - 3 / 4)

By taking LCM,

(8 + 1 / 4) / (8 - 3 / 4)

(9 / 4) / (5 / 4)

9 / 4 * 4 / 5

9 / 5 (4 and 4 cancel as they are in numerator and denominator)

1.8

Conclusion:

Therefore, 2 + cos ^ 2 A / 2 - sin ^ 2 A (tan A = √3) = 9 / 5 = 1.8

$<marquee> HOPE IT HELPS YOU...............XD$

Answer 2

Given:

tan A = √3 and 2 + cos ^ 2 A / 2 - sin ^ 2 A

To find:

The value of 2 + cos ^ 2 A / 2 - sin ^ 2 A

Solution:

tan A = √3

When A is 60° then, tan 60° = √3.

Therefore, A = 60°

Now, Substitute A = 60° in 2 + cos ^ 2 A / 2 - sin ^ 2 A

2 + cos ^ 2 A / 2 - sin ^ 2 A

2 + cos ^ 2 60° / 2 - sin ^ 2 60°

2 + (1 / 2)² / 2 - (√3 / 2)²

(2 + 1 / 4) / (2 - 3 / 4)

By taking LCM,

(8 + 1 / 4) / (8 - 3 / 4)

(9 / 4) / (5 / 4)

9 / 4 * 4 / 5

9 / 5 (4 and 4 cancel as they are in numerator and denominator)

1.8

Conclusion:

Therefore, 2 + cos ^ 2 A / 2 - sin ^ 2 A (tan A = √3) = 9 / 5 = 1.8