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Answer 1

Answer:

sin A / sin B = √3/2

⇒ sin²A / sin²B = 3/4

⇒ sin²A = 3/4 × sin²B -----(1)

cos A / cos B = √5/2

⇒ cos²A / cos²B = 5/4

⇒ cos²A = 5/4 cos²B  ----(2)

We know that sin²A + cos²A = 1 and hence on adding (1) and (2) we get :

⇒ 3/4 cos²B + 5/4 sin²B = 1

⇒ 3 cos²B + 5 sin²B = 4

⇒ 3 ( 1 - sin²B ) + 5 sin²B = 4

⇒ 3 - 3 sin²B + 5 sin²B = 4

⇒ 3 + 2 sin²B = 4

⇒ 2 sin²B = 4 - 3

⇒ 2 sin²B = 1

⇒ sin²B = 1/2

⇒ sin B = 1/√2

So we put the value of sin B to get :

3/4 cos²B + 5/4 sin²B = 1

⇒ 3/4 cos²B + 5/4 × 1/2 = 1

⇒ 3/4 cos²B = 1 - 5/8

⇒ 3/4 cos²B = ( 8 - 5 ) / 8

⇒ 3/4 cos²B = 3/8

⇒ cos²B = 1/2

⇒ cos B = 1/√2

We know that tan B = sin B / cos B

⇒ tan B = 1/√2 / 1/√2

⇒ tan B = 1

Dividing equation (1) and equation (2) we get :-

sin²A / cos²A / sin²B / cos²B = 3/4 / 5/4

⇒ tan²A / tan²B = 3/5

⇒ tan A / tan B = √3/√5

We already found out that tan B = 1 ,

tan A = √3/√5

tan A + tan B = √3/√5 + 1

⇒ tan A + tan B = ( √3 + √5 ) / √5

Rationalize further to get :

⇒ tan A + tan B = √5/√5 × ( √3 + √5 ) / √5

⇒ tan A + tan B = ( √15 + 5 ) / 5

Thus the value is ( √15 + 5 ) / 5 .

Answer 2

ANSWER

$\dfrac{\sqrt15+5}{5}$

Step By Step Explanation

Plz refer to the attachment for detailed answer given.

Remember

1)Sin A =$\dfrac{Perpendicular} {Hypotenuse}$

2)Cos A =$\dfrac{Base} {Hypotenuse}$

3)Tan A =$\dfrac {Perpendicular}{base}$

4)Sin A = $\dfrac{1}{Cosec A}$

5)Cos A =$\dfrac{1}{Sec A}$

6)Tan A =$\dfrac{1}{Cot A}$

7)$Sin^2A + Cos^2A$ = 1

8) $Sec^2A - Tan^2A$ = 1

9) $Cosec^2A - Cot^2A$ = 1

10)All values of trigonometric ratios or the trigonometric table are to be remembered.