Question

can you explain this​
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Answer 1

Step-by-step explanation:

Given :-

$4 cot A = 3$

To find :-

The value of $\frac{(sin A+ cos A)}{(sin A - cos A)}$

Solution :-

Given that

4 cot A = 3

=> cot A = 3/4 --------(1)

Now

$\frac{(sin A+ cos A)}{(sin A - cos A)}$

On dividing both numerator and denominator by sin A then

[(sin A+cos A)/(sin A)]/[(sin A-cos A)/(sin A)]

= [1+(cos A / sin A)] / [ 1- (cos A / sin A)]

= (1+cot A ) / (1-cot A)

= [1+(3/4)] / [1-(3/4)]

= [(4+3)/4]/ [ (4-3)/4]

= (7/4)/(1/4)

= (7/4)×(4/1)

= 7×1

= 7

Answer :-

The value of $\frac{(sin A+ cos A)}{(sin A - cos A)}$ is 7

Used formulae:-

♦ cot A = cos A / sin A

Answer 2

Answer:

Given that

4 cot A = 3

= cot A = 3/4 --------(1)

Now

$\frac{ \sin(a) + \cos(a) }{ \sin(a) - \cos(a) }$

On dividing both numerator and denominator by sin A then

[(sin A+cos A)/(sin A)]/[(sin A-cos A)/(sin A)]

= [1+(cos A / sin A)] / [ 1- (cos A / sin A)]

= (1+cot A ) / (1-cot A)

= [1+(3/4)] / [1-(3/4)]

= [(4+3)/4]/ [ (4-3)/4]

= (7/4)/(1/4)

= (7/4)×(4/1)

= 7×1

= 7