Question

if cos theta =5/13.Find 5 tan theta+12cot theta by
5 tan theta-12 cot theta.
​

Answer 1

${\underline {\underline {\huge {\sf {Provided \: Question: } }}}}$

If $\sf { cos \: \theta = \dfrac{5}{13} }$,Find $\sf { \dfrac{5 \: tan \: \theta +12 \: cot \: \theta}{5 \: tan \: \theta -12 \: cot \: \theta} }$

${\underline {\underline {\huge {\sf { Required \: Solution: } }}}}$

Given:

• $\sf { cos \: \theta = \dfrac{5}{13} }$

To calculate:

• $\sf { \dfrac{5 \: tan \: \theta +12 \: cot \: \theta}{5 \: tan \: \theta -12 \: cot \: \theta} }$

Calculation:

We know that,

• $\sf { cos \: \theta = \dfrac{Base}{Hypotenuse} }$

So,

• Base = 5
• Hypotenuse = 13

By Pythagoras property:

$\sf {⇢ {H}^{2} = {B}^{2} + {P}^{2} }$

$\sf {⇢ {13}^{2} = {5}^{2} + {P}^{2} }$

$\sf {⇢ 169 = 25 + {P}^{2} }$

$\sf {⇢ {P}^{2} = 169-25 }$

$\sf {⇢ {P}^{2} = 144 }$

$\sf {⇢ P = \sqrt{144} }$

$\boxed { \sf \pink{⇢ P = 12 }}$

Now we have,

• Base = 5
• Hypotenuse = 13
• Perpendicular = 12

We know that,

$\sf {⇢ tan \: \theta = \dfrac{Perpendicular}{Base} }$

${ \boxed { \sf {⇢ tan \: \theta = \dfrac{12}{5} }}}$

Also,

$\sf {⇢ cot \: \theta = \dfrac{Base}{Perpendicular} }$

${ \boxed { \sf {⇢ cot \: \theta = \dfrac{5}{12} }}}$

Substitute the values in the given expression:

$\sf {⇢ \dfrac{5 \: tan \: \theta +12 \: cot \: \theta}{5 \: tan \: \theta -12 \: cot \: \theta} }$

⠀⠀⠀

$\sf {⇢ \dfrac{ \Bigg(\cancel{5} \times \dfrac{12}{\cancel{5}} \Bigg) + \Bigg(\cancel{12} \times \dfrac{5}{\cancel{12}} \Bigg)}{\Bigg( \cancel{5} \times \dfrac{12}{\cancel{5} } \Bigg) -\Bigg(\cancel{12} \times \dfrac{5}{\cancel{12} } \Bigg)} }$

⠀⠀⠀

$\sf {⇢ \dfrac{12+ 5}{12-5} }$

⠀⠀⠀

$\boxed { \sf \pink{⇢ \dfrac{17}{7} }}$

____________________________________

Answer 2

Answer:

Let

△

A

B

C

be a right angled triangle where

∠

B

=

90

0

and

∠

C

=

θ

as shown in the above figure:

Now it is given that

cos

θ

=

5

13

and we know that, in a right angled triangle,

cos

θ

is equal to adjacent side over hypotenuse that is

cos

θ

=

A

d

j

a

c

e

n

t

s

i

d

e

H

y

p

o

t

e

n

u

s

e

, therefore, adjacent side

B

C

=

5

and hypotenuse

A

C

=

13

.

Now, using pythagoras theorem in

△

A

B

C

, we have

A

C

2

=

A

B

2

+

B

C

2

⇒

13

2

=

A

B

2

+

5

2

⇒

169

=

A

B

2

+

25

⇒

A

B

2

=

169

−

25

=

144

⇒

A

B

=

√

144

=

12

Therefore, the opposite side

A

B

=

12

.

We know that, in a right angled triangle,

tan

θ

is equal to opposite side over adjacent side that is

tan

θ

=

O

p

p

o

s

i

t

e

s

i

d

e

A

d

j

a

c

e

n

t

s

i

d

e

Here, we have opposite side

A

B

=

12

, adjacent side

B

C

=

5

and the hypotenuse

A

C

=

13

, therefore,

tan

θ

and

cot

θ

can be determined as follows:

tan

θ

=

O

p

p

o

s

i

t

e

s

i

d

e

A

d

j

a

c

e

n

t

s

i

d

e

=

A

B

B

C

=

12

5

cot

θ

=

1

tan

θ

=

1

12

5

=

1

×

5

12

=

5

12

Now, we find the following:

5

tan

θ

+

12

cot

θ

5

tan

θ

−

12

cot

θ

=

5

(

12

5

)

+

12

(

5

12

)

5

(

12

5

)

−

12

(

5

12

)

=

12

+

5

12

−

5

=

17

7

Hence,

5

tan

θ

+

12

cot

θ

5

tan

θ

−

12

cot

θ

=

17

7

.