Question

if (2sinθ-cosθ)/(cosθ+sinθ) =1 th3n th3 value of cotθ is

Answer 1

The value of cotθ is 1/2.

Step-by-step explanation:

In the question, it has been stated that (2sinθ-cosθ)/(cosθ+sinθ) equals to 1. We've been asked to find the value of cotθ by considering the given data.

In order to obtain the value of cotθ, let us write the given data properly, first.

$\twoheadrightarrow{ \sf{ \dfrac{2 \sin\theta - \cos \theta}{ \cos \theta + \sin \theta } = 1 }}$

Writing the value of L.H.S. in fractional form.

$\twoheadrightarrow{ \sf{ \dfrac{2 \sin\theta - \cos \theta}{ \cos \theta + \sin \theta } = \dfrac{1}{1} }}$

Cross multiplying the terms in the L.H.S. with those of in the R.H.S.

$\twoheadrightarrow{ \sf{1(2 \sin \theta - \cos \theta) = 1( \cos \theta + \sin \theta)}}$

Simplifying the terms in both the L.H.S. and the R.H.S.

$\twoheadrightarrow{ \sf{2 \sin \theta - \cos \theta = \cos \theta + \sin \theta}}$

Transposing the like terms to the L.H.S. and the R.H.S., appropriately.

$\twoheadrightarrow{ \sf{2 \sin \theta - \sin \theta = \cos \theta + \cos \theta}}$

Subtracting the terms in the L.H.S.

$\twoheadrightarrow{ \sf{ \sin \theta = \cos \theta + \cos \theta}}$

Adding the terms in the R.H.S.

$\twoheadrightarrow{ \sf{ \sin \theta = 2 \cos \theta}}$

Again transposing the like terms from the R.H.S. to the L.H.S.

$\twoheadrightarrow{ \sf{ \dfrac{ \sin \theta}{ \cos \theta} = 2 }}$

Since (sinθ)/(cosθ) equals tanθ, the equation can be written in the following way.

$\twoheadrightarrow{ \sf{ \tan \theta = 2}}$

By using the reciprocal Identity, the value of cotθ can be obtained as, cotθ equals to the reciprocal of tanθ.

$\twoheadrightarrow{ \sf{ \cot \theta = \dfrac{1}{2} }}$

And yeah, this is the required answer.

Therefore, cotθ equals to 1/2.

Answer 2

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$\maltese \; \; {\underline{\underline{\textsf{\textbf{Given that :}}}}}$

•  ${\underline{\boxed{\tt{ \dfrac{ (2\sin(\theta) - \cos(\theta))}{\cos(\theta) +\sin(\theta) } = 1}}}}$

$\maltese \; \; {\underline{\underline{\textsf{\textbf{To Find :}}}}}$

• The value of cot θ

$\maltese \; \; {\underline{\underline{\textsf{\textbf{Required Solution :}}}}}$

$\bigstar$ we know that the value of the expression (2sinθ-cosθ)/(cosθ+sinθ) equals  to the numerical result 1 , so now let's use suitablee ways to find out the value of cot θ from the given equation :

$\longrightarrow \sf \dfrac{2 \sin(\theta )- \cos(\theta) }{\sin(\theta )+ \cos(\theta)} = \dfrac{1}{1}$

Now let us consider the angle theta as alpha her and cross multiply the fraction which is in the from a/b = c/d in such way that a × d = c × b

$\longrightarrow \sf 1(2 \sin\alpha- \cos\alpha ) = 1 (\sin\alpha +\cos\alpha )$

$\longrightarrow \sf 2 \sin\alpha - \cos\alpha = \sin\alpha+ \cos\alpha$

As we have cross multiplied and simplified the terms now let's use the transposition method to solve the question

$\longrightarrow \sf 2 \sin\alpha - \sin\alpha = \cos\alpha + \cos\alpha$

$\longrightarrow \sf \sin\alpha = 2 \cos\alpha$

$\longrightarrow \sf \dfrac{\sin \alpha}{\cos \alpha} = 2$

We know that the value of sin/cos is equal to tan theta which is the reciprocal ratio of cot theta so, now we get the result

$\longrightarrow \sf \tan \theta = 2$

$\longrightarrow \sf \cot \theta = 1/2$

$\longrightarrow {\red{\underline{\underline{\rm{The \; value \; of \; \cot\theta \; equals \; to \;1/2 }}}}}$

$\maltese \; \; {\underline{\underline{\textsf{\textbf{Therefore :}}}}}$

• Cot θ equals to the numerical value 1/2

$\maltese \; \; {\underline{\underline{\textsf{\textbf{Additional information :}}}}}$

$\longrightarrow \sf \sin ( A + B ) = \sin A \cos B + \cos A \sin B$

$\longrightarrow \sf \sin ( A - B ) = \sin A \cos B - \cos A \sin B$

$\longrightarrow \sf \cos ( A + B ) = \cos A \cos B - \sin A \sin B$

$\longrightarrow \sf \cos ( A - B ) = \cos A \cos B + \sin A \sin B$

$\longrightarrow \sf \sin ( A + B ) +\sin ( A - B ) =2 \sin A \cos B$

$\longrightarrow \sf \sin ( A + B ) -\sin ( A - B ) =2 \cos A \sin B$

$\longrightarrow \sf \cos ( A + B ) +\cos ( A - B ) =2 \cos A \sin B$

$\longrightarrow \sf \cos ( A + B ) -\cos ( A - B ) =2 \sin A \sin B$

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