Question

given ,4cot A= 3,the value of $\implies \dfrac{sinA +cosA }{sinA-cosA}$is

a)7
b)2/11
c)1/2
d)1​

Answer 1

Answer :-

a) 7

Given :-

The expression -

→ 4cot A = 3

To Find :-

The value of -

$\sf{\dfrac{sinA+cosA}{sinA-cosA}}$

Solution :-

$\sf{4cot A = 3}$

$\sf{\longrightarrow cotA=\dfrac{3}{4}}$

$\sf{\longrightarrow \dfrac{cosA}{sinA}=\dfrac{3}{4}}$

Let us assume cosA = 3x and sinA = 4x.

$\sf{\dfrac{sinA+cosA}{sinA-cosA}=\dfrac{4x+3x}{4x-3x}}$

$\sf{\implies \dfrac{sinA+cosA}{sinA-cosA}=\dfrac{7x}{x} }$

$\sf{\implies \dfrac{sinA+cosA}{sinA-cosA}=7}$

Therefore, the answer is a) 7.

Answer 2

Given:-

4cotA = 3

To find:-

$\frac{sinA + cosA}{sinA - cosA}$=?

Solution:-

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

Now, Base = 3 and Perpendicular = 4

(Hypotenuse)² = (Perpendicular)² + (Base)²

(Hypotenuse)² = 3²+4² = 9 +16 = 25

Hypotenuse = √25 = 5

$sinA = \frac{4}{5} \\ cosA = \frac{3}{5}$

Now,

$= > \frac{sinA + cosA}{sinA - cosA} = \frac{ \frac{4}{5} + \frac{3}{5} }{ \frac{4}{5} - \frac{3}{5} }</p><p> = > \frac{ \frac{7}{5} }{ \frac{1}{5} } = \frac{7}{1}$

Hope its help uh❤