Question

If tanA = 4/3, what is the value of sinA + cosA / sinA - cosA

Answer 1

$Given \: tan A = \frac{4}{3} \: ---(1)$

$Value \:of \: \frac{sin A + cos A}{sin A - cos A}$

/* Dividing numerator and denominator by cosA ,we get */

$= \frac{\frac{(sin A + cos A)}{cosA}}{\frac{(sin A - cos A)}{cosA}}\\= \frac{\frac{sinA}{cosA} + \frac{cosA}{cosA}}{\frac{sinA}{cosA} - \frac{cosA}{sinA}}\\= \frac{tanA + 1 }{tanA - 1}$

$\boxed {\pink { Since, \frac{sinA}{cosA} = tan A }}$

$= \frac{\frac{4}{3} + 1}{\frac{4}{3} - 1 } \: [From \:(1)]$

$= \frac{\frac{4+3}{3}}{\frac{4-3}{3}}\\= \frac{\frac{7}{3}}{\frac{1}{3}} \\= \frac{7}{3} \times \frac{3}{1} \\= 7$

Therefore.,

$\red{Value \:of \: \frac{sin A + cos A}{sin A - cos A}}\green {= 7}$

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Answer 2

Answer:

GiventanA=

3

4

−−−(1)

Value \:of \: \frac{sin A + cos A}{sin A - cos A} Valueof

sinA−cosA

sinA+cosA

/* Dividing numerator and denominator by cosA ,we get */

\begin{gathered} = \frac{\frac{(sin A + cos A)}{cosA}}{\frac{(sin A - cos A)}{cosA}}\\= \frac{\frac{sinA}{cosA} + \frac{cosA}{cosA}}{\frac{sinA}{cosA} - \frac{cosA}{sinA}}\\= \frac{tanA + 1 }{tanA - 1} \end{gathered}

=

cosA

(sinA−cosA)

cosA

(sinA+cosA)

=

cosA

sinA

−

sinA

cosA

cosA

sinA

+

cosA

cosA

=

tanA−1

tanA+1

\boxed {\pink { Since, \frac{sinA}{cosA} = tan A }}

Since,

cosA

sinA

=tanA

= \frac{\frac{4}{3} + 1}{\frac{4}{3} - 1 } \: [From \:(1)] =

3

4

−1

3

4

+1

[From(1)]

\begin{gathered} = \frac{\frac{4+3}{3}}{\frac{4-3}{3}}\\= \frac{\frac{7}{3}}{\frac{1}{3}} \\= \frac{7}{3} \times \frac{3}{1} \\= 7 \end{gathered}

=

3

4−3

3

4+3

=

3

1

3

7

=

3

7

×

1

3

=7

Therefore.,

\red{Value \:of \: \frac{sin A + cos A}{sin A - cos A}}\green {= 7} Valueof

sinA−cosA

sinA+cosA

=7