Question

if cos A = 3/5 evaluate 5 sin A + 3sec A - 3tan A / 4 cot A + 4 cosec A +. 5 cos A​

Answer 1

$\bf \large \underline{Answer:}$

Given :

cos A = 3/5

To Find:

5 sin A + 3sec A - 3tan A / 4 cot A + 4 cosec A +5 cos A

Solution :

First step find the value for sin A, sec A , tan A, cot A , cosec A.

we know that,

cos A = 3/5 = adjacent side / hypotenuse side ,

now find the opposite side,

BC ² = AC² - AB²

BC²= 5²-3²

BC² = 25-9

BC²=16

BC =√16

BC = 4 cm(opposite side)

Now ,

★ sin A

sin A = opposite side/ hypotenuse side

sin A = BC /AC = 4/5

★sec A

sec A is reciprocal of cosA

sec A = 1/ cosA = 1/3/5 = 5/3

★tan A = opposite side/ adjacent side

•°• tan A = 4/3

★cot A is reciprocal of tan A

•°• cot A = 3/4

★ cosec A is reciprocal of sinA

•°•cosec A = 5/4

Second step , substitute all the value in the given form :

$: \implies \tt \pink{\frac{5 \: sin \:A + 3 \: sec \: A - 3 \: tan \:A }{4 \: cot \:A + 4 \: cosec \:A + 5 \: cos \: A} }$

$: \implies \tt \pink{ \frac{5( \frac{4}{5}) + 3( \frac{5}{3}) - 3( \frac{4}{3} ) }{4( \frac{3}{4} ) + 4( \frac{5}{4}) + 5( \frac{3}{5}) } }$

$: \implies \tt \pink{ \frac{ \cancel5( \frac{4}{ \cancel5}) + \cancel3( \frac{5}{\cancel3}) - \cancel3( \frac{4}{\cancel3} ) }{\cancel4( \frac{3}{\cancel4} ) + \cancel4( \frac{5}{\cancel4}) + \cancel5( \frac{3}{\cancel5}) } }$

$: \implies \tt \pink{ \frac{4 + 5 - 4}{3 + 5 + 3}}$

$: \implies \tt \pink{ \frac{5}{11}}$

$\therefore{ \boxed{\tt {\purple{\frac{5 \: sin \:A + 3 \: sec \: A - 3 \: tan \:A }{4 \: cot \:A + 4 \: cosec \:A + 5 \: cos \: A } = \frac{5}{11} } }}}$

Refer the attachment for the image !!

Answer 2

Step-by-step explanation:

\bf \large \underline{Answer:}

Answer:

Given :

cos A = 3/5

To Find:

5 sin A + 3sec A - 3tan A / 4 cot A + 4 cosec A +5 cos A

Solution :

First step find the value for sin A, sec A , tan A, cot A , cosec A.

we know that,

cos A = 3/5 = adjacent side / hypotenuse side ,

now find the opposite side,

BC ² = AC² - AB²

BC²= 5²-3²

BC² = 25-9

BC²=16

BC =√16

BC = 4 cm(opposite side)

Now ,

★ sin A

sin A = opposite side/ hypotenuse side

sin A = BC /AC = 4/5

★sec A

sec A is reciprocal of cosA

sec A = 1/ cosA = 1/3/5 = 5/3

★tan A = opposite side/ adjacent side

•°• tan A = 4/3

★cot A is reciprocal of tan A

•°• cot A = 3/4

★ cosec A is reciprocal of sinA

•°•cosec A = 5/4

Second step , substitute all the value in the given form :

\begin{gathered} : \implies \tt \pink{\frac{5 \: sin \:A + 3 \: sec \: A - 3 \: tan \:A }{4 \: cot \:A + 4 \: cosec \:A + 5 \: cos \: A} } \\ \end{gathered}

:⟹

4cotA+4cosecA+5cosA

5sinA+3secA−3tanA

\begin{gathered} : \implies \tt \pink{ \frac{5( \frac{4}{5}) + 3( \frac{5}{3}) - 3( \frac{4}{3} ) }{4( \frac{3}{4} ) + 4( \frac{5}{4}) + 5( \frac{3}{5}) } } \\ \end{gathered}

:⟹

4(

4

3

)+4(

4

5

)+5(

5

3

)

5(

5

4

)+3(

3

5

)−3(

3

4

)

\begin{gathered} : \implies \tt \pink{ \frac{ \cancel5( \frac{4}{ \cancel5}) + \cancel3( \frac{5}{\cancel3}) - \cancel3( \frac{4}{\cancel3} ) }{\cancel4( \frac{3}{\cancel4} ) + \cancel4( \frac{5}{\cancel4}) + \cancel5( \frac{3}{\cancel5}) } } \\ \end{gathered}

:⟹

4

(

4

3

)+

4

(

4

5

)+

5

(

5

3

)

5

(

5

4

)+

3

(

3

5

)−

3

(

3

4

)

\begin{gathered} : \implies \tt \pink{ \frac{4 + 5 - 4}{3 + 5 + 3}} \\ \end{gathered}

:⟹

3+5+3

4+5−4

\begin{gathered} : \implies \tt \pink{ \frac{5}{11}} \\ \end{gathered}

:⟹

11

5

\begin{gathered} \therefore{ \boxed{\tt {\purple{\frac{5 \: sin \:A + 3 \: sec \: A - 3 \: tan \:A }{4 \: cot \:A + 4 \: cosec \:A + 5 \: cos \: A } = \frac{5}{11} } }}} \\ \end{gathered}

∴

4cotA+4cosecA+5cosA

5sinA+3secA−3tanA

=

11

5

Refer the attachment for the image !!