Question

if cot theta = 5/4 then find the values of 5 sin theta + 3 cos theta / 5 sin theta - 2 cos theta

Answer 1

➝Correct Question :

If $cot\:\theta = \dfrac{3}{4}$, then find the value of:

$5sin\:\theta + \dfrac{3cos\:\theta}{5sin\:\theta} - 2cos\:theta$

➝ Find :

The value of :

$\boxed{\mathtt{5sin\:\theta + \dfrac{3cos\:\theta}{5sin\:\theta} - 2cos\:theta}}$

➝ Given :

The value of $cot\:\theta$

$\rightarrow cot\:\theta = \dfrac{3}{4}$

➝ We Know :

The trigonometrical identity of

• $cot\:\theta = \dfrac{b}{p}$

• $sin\:\theta = \dfrac{p}{h}$

• $cos\:\theta = \dfrac{b}{h}$

Where,

• b = base of the triangle
• p = height of triangle
• h = hypotenuse of the triangle

Pythagoras theorem :

$\boxed{\mathtt{h^{2} = p^{2} + b^{2}}}$

Where,

• h = Hypotenuse
• p = height
• b = base

➝ Concept :

According to the given information,that $cot\:\theta = \dfrac{3}{4}$, and the identity that $cot\:\theta = \dfrac{b}{p}$ ,we got the base and height of the the triangle ..i.e,

• Height = 4 units
• base = 3 units

From the above information of base and height , we can find the value of Hypotenuse by the Pythagoras theorem..

Pythagoras theorem :

$\mathtt{h^{2} = p^{2} + b^{2}}$

By using it ,and putting the value of height and base in the formula ,we get :

$\mathtt{\Rightarrow h^{2} = 4^{2} + 3^{2}}$

$\mathtt{\Rightarrow h = \sqrt{4^{2} + 3^{2}}}$

$\mathtt{\Rightarrow h = \sqrt{16 + 9}}$

$\mathtt{\Rightarrow h = \sqrt{25}}$

$\mathtt{\Rightarrow h = 5 units}$

Hence ,the hypotenuse of the triangle is 5 units.

By putting the value of different trigonometric identities , we can find the value of the given Equation :

• $sin\:\theta = \dfrac{4}{5}$

(As $sin\:\theta = \dfrac{p}{h}$)

• $cos\:\theta = \dfrac{3}{5}$

(As $cos\:\theta = \dfrac{b}{h}$)

➝ Solution :

We now know that :

• $sin\:\theta = \dfrac{4}{5}$
• $cos\:\theta = \dfrac{3}{5}$

Putting the value in the Equation ,

$5sin\:\theta + \dfrac{3cos\:\theta}{5sin\:\theta} - 2cos\:theta$

we get :

$\mathtt{\Rightarrow 5 \times \dfrac{4}{5} + \dfrac{3 \times \dfrac{3}{5}}{5} \times \dfrac{4}{5} - 2 \times \dfrac{3}{5}}$

$\mathtt{\Rightarrow \cancel{5} \times \dfrac{4}{\cancel{5}} + \dfrac{3 \times \dfrac{3}{5}}{\cancel{5} \times \dfrac{4}{\cancel{5}}} - 2 \times \dfrac{3}{5}}$

$\mathtt{\Rightarrow 4 + \dfrac{\dfrac{9}{5}}{4} - \dfrac{6}{5}}$

$\mathtt{\Rightarrow 4 + \dfrac{9}{5} \times 4 - \dfrac{6}{5}}$

$\mathtt{\Rightarrow 4 + \dfrac{36}{5} - \dfrac{6}{5}}$

$\mathtt{\Rightarrow \dfrac{20 + 36 - 6}{5}}$

$\mathtt{\Rightarrow \dfrac{56 - 6}{5}}$

$\mathtt{\Rightarrow \dfrac{50}{5}}$

$\mathtt{\Rightarrow \dfrac{\cancel{50}}{\cancel{5}}}$

$\mathtt{\Rightarrow 10}$

Hence ,the value of $5sin\:\theta + \dfrac{3cos\:\theta}{5sin\:\theta} - 2cos\:theta$ is $\mathtt{10}$

➝ Extra Information :

Some Trigonometrical identities :

• $sin^{2}\theta + cos^{2}\theta = 1$

• $sec^{2}\theta - tan^{2}\theta = 1$

• $cosec^{2}\theta - cot^{2}\theta = 1$

• $sin(A + B) = sinAcosB + cosAsinB$

• $sin(A - B) = sinAcosB - cosAsinB$

• $cos(A + B) = cosAcosB - sinAsinB$

• $cos(A - B) = cosAcosB + sinAsinB$

Answer 2

Answer:

[tex]

➝Correct Question :

If cot\:\theta = \dfrac{3}{4}cotθ=

4

3

, then find the value of:

5sin\:\theta + \dfrac{3cos\:\theta}{5sin\:\theta} - 2cos\:theta5sinθ+

5sinθ

3cosθ

−2costheta

➝ Find :

The value of :

\boxed{\mathtt{5sin\:\theta + \dfrac{3cos\:\theta}{5sin\:\theta} - 2cos\:theta}}

5sinθ+

5sinθ

3cosθ

−2costheta

➝ Given :

The value of cot\:\thetacotθ

\rightarrow cot\:\theta = \dfrac{3}{4}→cotθ=

4

3

➝ We Know :

The trigonometrical identity of

cot\:\theta = \dfrac{b}{p}cotθ=

p

b

sin\:\theta = \dfrac{p}{h}sinθ=

h

p

cos\:\theta = \dfrac{b}{h}cosθ=

h

b

Where,

b = base of the triangle

p = height of triangle

h = hypotenuse of the triangle

Pythagoras theorem :

\boxed{\mathtt{h^{2} = p^{2} + b^{2}}}

h

2

=p

2

+b

2

Where,

h = Hypotenuse

p = height

b = base

➝ Concept :

According to the given information,that cot\:\theta = \dfrac{3}{4}cotθ=

4

3

, and the identity that cot\:\theta = \dfrac{b}{p}cotθ=

p

b

,we got the base and height of the the triangle ..i.e,

Height = 4 units

base = 3 units

From the above information of base and height , we can find the value of Hypotenuse by the Pythagoras theorem..

Pythagoras theorem :

\mathtt{h^{2} = p^{2} + b^{2}}h

2

=p

2

+b

2

By using it ,and putting the value of height and base in the formula ,we get :

\mathtt{\Rightarrow h^{2} = 4^{2} + 3^{2}}⇒h

2

=4

2

+3

2

\mathtt{\Rightarrow h = \sqrt{4^{2} + 3^{2}}}⇒h=

4

2

+3

2

\mathtt{\Rightarrow h = \sqrt{16 + 9}}⇒h=

16+9

\mathtt{\Rightarrow h = \sqrt{25}}⇒h=

25

\mathtt{\Rightarrow h = 5 units}⇒h=5units

Hence ,the hypotenuse of the triangle is 5 units.

By putting the value of different trigonometric identities , we can find the value of the given Equation :

sin\:\theta = \dfrac{4}{5}sinθ=

5

4

(As sin\:\theta = \dfrac{p}{h}sinθ=

h

p

)

cos\:\theta = \dfrac{3}{5}cosθ=

5

3

(As cos\:\theta = \dfrac{b}{h}cosθ=

h

b

)

➝ Solution :

We now know that :

sin\:\theta = \dfrac{4}{5}sinθ=

5

4

cos\:\theta = \dfrac{3}{5}cosθ=

5

3

Putting the value in the Equation ,

5sin\:\theta + \dfrac{3cos\:\theta}{5sin\:\theta} - 2cos\:theta5sinθ+

5sinθ

3cosθ

−2costheta

we get :

\mathtt{\Rightarrow 5 \times \dfrac{4}{5} + \dfrac{3 \times \dfrac{3}{5}}{5} \times \dfrac{4}{5} - 2 \times \dfrac{3}{5}}⇒5×

5

4

+

5

3×

5

3

×

5

4

−2×

5

3

\mathtt{\Rightarrow \cancel{5} \times \dfrac{4}{\cancel{5}} + \dfrac{3 \times \dfrac{3}{5}}{\cancel{5} \times \dfrac{4}{\cancel{5}}} - 2 \times \dfrac{3}{5}}⇒

5

×

5

4

+

5

×

5

4

3×

5

3

−2×

5

3

\mathtt{\Rightarrow 4 + \dfrac{\dfrac{9}{5}}{4} - \dfrac{6}{5}}⇒4+

4

5

9

−

5

6

\mathtt{\Rightarrow 4 + \dfrac{9}{5} \times 4 - \dfrac{6}{5}}⇒4+

5

9

×4−

5

6

\mathtt{\Rightarrow 4 + \dfrac{36}{5} - \dfrac{6}{5}}⇒4+

5

36

−

5

6

\mathtt{\Rightarrow \dfrac{20 + 36 - 6}{5}}⇒

5

20+36−6

\mathtt{\Rightarrow \dfrac{56 - 6}{5}}⇒

5

56−6

\mathtt{\Rightarrow \dfrac{50}{5}}⇒

5

50

\mathtt{\Rightarrow \dfrac{\cancel{50}}{\cancel{5}}}⇒

5

50

\mathtt{\Rightarrow 10}⇒10

Hence ,the value of 5sin\:\theta + \dfrac{3cos\:\theta}{5sin\:\theta} - 2cos\:theta5sinθ+

5sinθ

3cosθ

−2costheta is \mathtt{10}10

➝ Extra Information :

Some Trigonometrical identities :

sin^{2}\theta + cos^{2}\theta = 1sin

2

θ+cos

2

θ=1

sec^{2}\theta - tan^{2}\theta = 1sec

2

θ−tan

2

θ=1

cosec^{2}\theta - cot^{2}\theta = 1cosec

2

θ−cot

2

θ=1

sin(A + B) = sinAcosB + cosAsinBsin(A+B)=sinAcosB+cosAsinB

sin(A - B) = sinAcosB - cosAsinBsin(A−B)=sinAcosB−cosAsinB

cos(A + B) = cosAcosB - sinAsinBcos(A+B)=cosAcosB−sinAsinB

cos(A - B) = cosAcosB + sinAsinBcos(A−B)=cosAcosB+sinAsinB

[/tex]