Question

[tex]\large\underline\bold{ANSWER \:FAST\red{\huge{\checkmark}}}[/tex


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Answer 1

Given:-

• sinA: cosA = 3:4

To Find:-

secA + cosecA = ?

Solution:-

$tanA = \frac{3}{4}$

Now , Perpendicular = 3 and base = 4

(Hypotenuse)² = (perpendicular)²+(base)²

(Hypotenuse)² = 3²+4² = 9 + 16 = 25

Hypotenuse = √25 = 5

$secA = \frac{5}{4} \\ coscecA = \frac{5}{3}$

Now, secA+cosecA = $\frac{5}{4} + \frac{5}{3} = \frac{35}{12}$

Hope its help uh❤

Answer 2

Answer :-

• 35/12

Given :-

• sinA : cosA = 3 : 4

To Find :-

• secA + cosecA

Solution :-

Here,

• sinA/cosA = 3/4
• tanA = 3/4

Perpendicular = 3

base = 4

As we know that

h² = p² + b²

Where,

• h = hypotenuse
• p = perpendicular
• b = base

Put the values in the formula

h² = (3)² + (4)²

→ h² = 9 + 16

→ h² = 25

→ h = √25

→ h = 5

secA = h/b = 5/4

cosecA = h/p = 5/3

secA + cosecA

→ 5/4 + 5/3

→ 35/12

Hence, secA + cosecA = 35/12.