If cot theta = 5/4 then find the values of 5 sin theta + 3 cos theta /

1.	If cot theta = 5/4 then find the values of 5 sin theta + 3 cos theta / 5 sin theta - 2 cos theta
Answer» ➝Correct Question : If $cot\:\theta = \dfrac{3}{<klux>4</klux>}$ , 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 : $sin^{2}\theta + cos^{2}\theta = 1$ $sec^{2}\theta - tan^{2}\theta = 1$ $cosec^{2}\theta - cot^{2}\theta = 1$ $sin(A + B) = sinAcosB + cosAsinB$ $sin(A - B) = sinAcosB - cosAsinB$ $cos(A + B) = cosAcosB - sinAsinB$ $cos(A - B) = cosAcosB + sinAsinB$

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

Answer»

➝Correct Question :

If $cot\:\theta = \dfrac{3}{<klux>4</klux>}$ , 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 :

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

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