Question

If 1 + cos² theta = 3 sin theta cos theta, then find the value of cot theta ....​

Answer 1

★ Concept :-

Here the concept of Trigonometry and Quadratic Equations have been used. We see that we are given a trigonometric equation. Now we have to find a required value. How can we do that ? A simple trick to do that is firstly divide all the terms in LHS and RHS by a common term. And then reduce them to find answer. We shall get a quadratic equation, which can be easily solved by Splitting the middle term.

Let's do it !!

___________________________________________

★ Formula Used :-

$\\\;\boxed{\sf{\pink{cosec\theta\;=\;\bf{\dfrac{1}{\sin\theta}}}}}$

$\\\;\boxed{\sf{\pink{\cot\theta\;=\;\bf{\dfrac{\cos\theta}{\sin\theta}}}}}$

$\\\;\boxed{\sf{\pink{\cosec^{2}\theta\;=\;\bf{1\;+\;\cot^{2}\theta}}}}$

___________________________________________

★ Solution :-

Given,

✒ 1 + cos² θ = 3 sin θ cos θ

Firstly let's divide all the terms of LHS and RHS by sin² θ . Then we get ,

$\\\;\sf{\rightarrow\;\;\dfrac{1}{\sin^{2}\theta}\;+\;\dfrac{\cos^{2}\theta}{\sin^{2}\theta}\;=\;\bf{\dfrac{3\sin\theta\:\cos\theta}{\sin^{2}\theta}}}$

We already know that,

$\\\;\tt{\leadsto\;\;cosec\theta\;=\;\dfrac{1}{\sin\theta}}$

By applying this in the above equation, we get

$\\\;\sf{\rightarrow\;\;\bigg(\dfrac{1}{\sin\theta}\bigg)^{2}\;+\;\dfrac{\cos^{2}\theta}{\sin^{2}\theta}\;=\;\bf{\dfrac{3\sin\theta\:\cos\theta}{\sin^{2}\theta}}}$

Since, 1² = 1,

$\\\;\sf{\rightarrow\;\;cosec^{2}\theta\;+\;\dfrac{\cos^{2}\theta}{\sin^{2}\theta}\;=\;\bf{\dfrac{3\sin\theta\:\cos\theta}{\sin^{2}\theta}}}$

Cancelling sin θ from RHS, we get

$\\\;\sf{\rightarrow\;\;cosec^{2}\theta\;+\;\dfrac{\cos^{2}\theta}{\sin^{2}\theta}\;=\;\bf{\dfrac{3\cos\theta}{\sin\theta}}}$

This can be written as,

$\\\;\sf{\rightarrow\;\;cosec^{2}\theta\;+\;\bigg(\dfrac{\cos\theta}{\sin\theta}\bigg)^{2}\;=\;\bf{\dfrac{3\cos\theta}{\sin\theta}}}$

Also, we know that

$\\\;\tt{\leadsto\;\;\cot\theta\;=\;\dfrac{\cos\theta}{\sin\theta}}$

By applying this in the above equation we get,

$\\\;\sf{\rightarrow\;\;cosec^{2}\theta\;+\;\bigg(\cot\theta\bigg)^{2}\;=\;\bf{3\cot\theta}}$

$\\\;\sf{\rightarrow\;\;\red{cosec^{2}\theta\;+\;\cot^{2}\theta\;=\;\bf{3\cot\theta}}}$

We also know that,

$\\\;\tt{\leadsto\;\;cosec^{2}\theta\;=\;1\;+\;\cot^{2}\theta}$

By applying this in the given equation, we get

$\\\;\sf{\rightarrow\;\;1\;+\;\cot^{2}\theta\;+\;\cot^{2}\theta\;=\;\bf{3\cot\theta}}$

$\\\;\sf{\rightarrow\;\;1\;+\;2\cot^{2}\theta\;=\;\bf{3\cot\theta}}$

$\\\;\sf{\rightarrow\;\;1\;+\;2\cot^{2}\theta\;-\;3\cot\theta\;=\;\bf{0}}$

By rearranging, we get

$\\\;\sf{\rightarrow\;\;\green{2\cot^{2}\theta\;-\;3\cot\theta\;+\;1\;=\;\bf{0}}}$

Here we get a quadratic equation.

• Let cot θ = x

By applying this, we get

$\\\;\sf{\rightarrow\;\;2x^{2}\;-\;3x\;+\;1\;=\;\bf{0}}$

Now we can use the method of splitting the middle term to find our answer.

$\\\;\sf{\Longrightarrow\;\;2x^{2}\;-\;2x\;-\;x\;+\;1\;=\;\bf{0}}$

(Since 2 × 1 = 2)

By taking the terms common, we get

$\\\;\sf{\Longrightarrow\;\;2x(x\;-\;1)\;-\;1(x\;-\;1)\;=\;\bf{0}}$

$\\\;\sf{\Longrightarrow\;\;\orange{(2x\;-\;1)(x\;-\;1)\;=\;\bf{0}}}$

Here since (2x - 1) and (x - 1) are being multiplied, so either (2x - 1) = 0 or (x - 1) = 0 .

Then,

✒ (2x - 1) = 0 or (x - 1) = 0

✒ 2x = 1 or x = 1

✒ x = ½ or x = 1

✒ x = ½ , 1

We already know that, x = cot θ . So,

✒ cot θ = ½ , 1

This is the required answer. So,

$\\\;\underline{\boxed{\tt{Required\;\:value\;\:of\;\:\cot\theta\;=\;\bf{\purple{\dfrac{1}{2}\:,\:1}}}}}$

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★ More to know :-

$\\\;\tt{\leadsto\;\;\sin^{2}\theta\;+\;\cos^{2}\theta\;=\;1}$

$\\\;\tt{\leadsto\;\;\sec^{2}\theta\;=\;1\;+\;\cot^{2}\theta}$

$\\\;\tt{\leadsto\;\;\sec\theta\;=\;\dfrac{1}{\cos\theta}}$

$\\\;\tt{\leadsto\;\;\tan\theta\;=\;\dfrac{\sin\theta}{\cos\theta}}$

Answer 2

Answer:

★ Concept :-

Here the concept of Trigonometry and Quadratic Equations have been used. We see that we are given a trigonometric equation. Now we have to find a required value. How can we do that ? A simple trick to do that is firstly divide all the terms in LHS and RHS by a common term. And then reduce them to find answer. We shall get a quadratic equation, which can be easily solved by Splitting the middle term.

Let's do it !!

___________________________________________

★ Formula Used :-

$\begin{gathered}\\\;\boxed{\sf{\pink{cosec\theta\;=\;\bf{\dfrac{1}{\sin\theta}}}}}\end{gathered}$

$\begin{gathered}\\\;\boxed{\sf{\pink{\cot\theta\;=\;\bf{\dfrac{\cos\theta}{\sin\theta}}}}}\end{gathered}$

$\begin{gathered}\\\;\boxed{\sf{\pink{\cosec^{2}\theta\;=\;\bf{1\;+\;\cot^{2}\theta}}}}\end{gathered}$

___________________________________________

★ Solution :-

Given,

✒ 1 + cos² θ = 3 sin θ cos θ

Firstly let's divide all the terms of LHS and RHS by sin² θ . Then we get ,

$\begin{gathered}\\\;\sf{\rightarrow\;\;\dfrac{1}{\sin^{2}\theta}\;+\;\dfrac{\cos^{2}\theta}{\sin^{2}\theta}\;=\;\bf{\dfrac{3\sin\theta\:\cos\theta}{\sin^{2}\theta}}}\end{gathered}$

We already know that,

$\begin{gathered}\\\;\tt{\leadsto\;\;cosec\theta\;=\;\dfrac{1}{\sin\theta}}\end{gathered}$

By applying this in the above equation, we get

$\begin{gathered}\\\;\sf{\rightarrow\;\;\bigg(\dfrac{1}{\sin\theta}\bigg)^{2}\;+\;\dfrac{\cos^{2}\theta}{\sin^{2}\theta}\;=\;\bf{\dfrac{3\sin\theta\:\cos\theta}{\sin^{2}\theta}}}\end{gathered}$

Since, 1² = 1,

$\begin{gathered}\\\;\sf{\rightarrow\;\;cosec^{2}\theta\;+\;\dfrac{\cos^{2}\theta}{\sin^{2}\theta}\;=\;\bf{\dfrac{3\sin\theta\:\cos\theta}{\sin^{2}\theta}}}\end{gathered}$

Cancelling sin θ from RHS, we get

$\begin{gathered}\\\;\sf{\rightarrow\;\;cosec^{2}\theta\;+\;\dfrac{\cos^{2}\theta}{\sin^{2}\theta}\;=\;\bf{\dfrac{3\cos\theta}{\sin\theta}}}\end{gathered}$

This can be written as,

$\begin{gathered}\\\;\sf{\rightarrow\;\;cosec^{2}\theta\;+\;\bigg(\dfrac{\cos\theta}{\sin\theta}\bigg)^{2}\;=\;\bf{\dfrac{3\cos\theta}{\sin\theta}}}\end{gathered}$

Also, we know that

$\begin{gathered}\\\;\tt{\leadsto\;\;\cot\theta\;=\;\dfrac{\cos\theta}{\sin\theta}}\end{gathered}$

By applying this in the above equation we get,

$\begin{gathered}\\\;\sf{\rightarrow\;\;cosec^{2}\theta\;+\;\bigg(\cot\theta\bigg)^{2}\;=\;\bf{3\cot\theta}}\end{gathered}$

$\begin{gathered}\\\;\sf{\rightarrow\;\;\red{cosec^{2}\theta\;+\;\cot^{2}\theta\;=\;\bf{3\cot\theta}}}\end{gathered}$

We also know that,

$\begin{gathered}\\\;\tt{\leadsto\;\;cosec^{2}\theta\;=\;1\;+\;\cot^{2}\theta}\end{gathered}$

By applying this in the given equation, we get

$\begin{gathered}\\\;\sf{\rightarrow\;\;1\;+\;\cot^{2}\theta\;+\;\cot^{2}\theta\;=\;\bf{3\cot\theta}}\end{gathered}$

$\begin{gathered}\\\;\sf{\rightarrow\;\;1\;+\;2\cot^{2}\theta\;=\;\bf{3\cot\theta}}\end{gathered}$

$\begin{gathered}\\\;\sf{\rightarrow\;\;1\;+\;2\cot^{2}\theta\;-\;3\cot\theta\;=\;\bf{0}}\end{gathered}$

By rearranging, we get

$\begin{gathered}\\\;\sf{\rightarrow\;\;\green{2\cot^{2}\theta\;-\;3\cot\theta\;+\;1\;=\;\bf{0}}}\end{gathered}$

Here we get a quadratic equation.

Let cot θ = x

By applying this, we get

$\begin{gathered}\\\;\sf{\rightarrow\;\;2x^{2}\;-\;3x\;+\;1\;=\;\bf{0}}\end{gathered}$

Now we can use the method of splitting the middle term to find our answer.

$\begin{gathered}\\\;\sf{\Longrightarrow\;\;2x^{2}\;-\;2x\;-\;x\;+\;1\;=\;\bf{0}}\end{gathered}$

(Since 2 × 1 = 2)

By taking the terms common, we get

$\begin{gathered}\\\;\sf{\Longrightarrow\;\;2x(x\;-\;1)\;-\;1(x\;-\;1)\;=\;\bf{0}}\end{gathered}$

$\begin{gathered}\\\;\sf{\Longrightarrow\;\;\orange{(2x\;-\;1)(x\;-\;1)\;=\;\bf{0}}}\end{gathered}$

Here since (2x - 1) and (x - 1) are being multiplied, so either (2x - 1) = 0 or (x - 1) = 0 .

Then,

✒ (2x - 1) = 0 or (x - 1) = 0

✒ 2x = 1 or x = 1

✒ x = ½ or x = 1

✒ x = ½ , 1

We already know that, x = cot θ . So,

✒ cot θ = ½ , 1

This is the required answer. So,

$\begin{gathered}\\\;\underline{\boxed{\tt{Required\;\:value\;\:of\;\:\cot\theta\;=\;\bf{\purple{\dfrac{1}{2}\:,\:1}}}}}\end{gathered}$

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★ More to know :-

$\begin{gathered}\\\;\tt{\leadsto\;\;\sin^{2}\theta\;+\;\cos^{2}\theta\;=\;1}\end{gathered}$

$\begin{gathered}\\\;\tt{\leadsto\;\;\sec^{2}\theta\;=\;1\;+\;\cot^{2}\theta}\end{gathered}$

$\begin{gathered}\\\;\tt{\leadsto\;\;\sec\theta\;=\;\dfrac{1}{\cos\theta}}\end{gathered}$

$\begin{gathered}\\\;\tt{\leadsto\;\;\tan\theta\;=\;\dfrac{\sin\theta}{\cos\theta}}\end{gathered}$