
Find the value of $ \sqrt {4{{\cos }^4}\theta + {{\sin }^2}2\theta } + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$, if \[\dfrac{\pi }{2} < \theta < \dfrac{{3\pi }}{2}\].
A) $2\sin \theta $
B) $ - 2\sin \theta $
C) $2\cot \theta $
D) $ - 2\cot \theta $
Answer
576.3k+ views
Hint: we can solve this problem by the simplifying given term.
Here we can use some trigonometric formulas.
1) $\sin 2\theta = 2\sin \theta \cos \theta $
2) $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
3) $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
4) ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Complete step by step answer:
1) Let us simplify the given term.
We have,
$ = \sqrt {4{{\cos }^4}\theta + {{\sin }^2}2\theta } + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$
Let us simplify first the terms in the square root.
We can use 1) formula.
We get,
$ \Rightarrow \sqrt {4{{\cos }^4}\theta + {{\left( {2\sin \theta \cos \theta } \right)}^2}} + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$
Simplify the remaining term.
Let us use 2) formula.
Now we get,
$ \Rightarrow \sqrt {4{{\cos }^4}\theta + {{\left( {2\sin \theta \cos \theta } \right)}^2}} + 4\cot \theta {\left[ {\cos \dfrac{\pi }{4}\cos \dfrac{\theta }{2} + \sin \dfrac{\pi }{4}\sin \dfrac{\theta }{2}} \right]^2}$
Now, we can put the values of cos and sin.
As we know the values of, $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and$\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$.
Substitute these values.
$ \Rightarrow \sqrt {4{{\cos }^4}\theta + {{\left( {2\sin \theta \cos \theta } \right)}^2}} + 4\cot \theta {\left[ {\dfrac{1}{{\sqrt 2 }}\cos \dfrac{\theta }{2} + \dfrac{1}{{\sqrt 2 }}\sin \dfrac{\theta }{2}} \right]^2}$
Consider the term which is under the square root and is in the bracket.
Take the square of that term.
See, $\dfrac{1}{{\sqrt 2 }}$ is in both the terms.
So, let us write it outside the bracket as a common term.
So we get,
$ \Rightarrow \sqrt {4{{\cos }^4}\theta + \left( {4{{\sin }^2}\theta {{\cos }^2}\theta } \right)} + \dfrac{4}{{{{\left( {\sqrt 2 } \right)}^2}}}\cot \theta {\left[ {\cos \left( {\dfrac{\theta }{2}} \right) + \sin \left( {\dfrac{\theta }{2}} \right)} \right]^2}$
Consider the term which is under the square root.
Let us take common terms outside the bracket.
So we get,
$ \Rightarrow \sqrt {4{{\cos }^2}\theta \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)} + \dfrac{4}{{{{\left( {\sqrt 2 } \right)}^2}}}\cot \theta {\left[ {\cos \left( {\dfrac{\theta }{2}} \right) + \sin \left( {\dfrac{\theta }{2}} \right)} \right]^2}$
While removing the square root so, the value which is in the square root will be in the mod.
After that, we will use the formula to simplify the bracket:${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.
$ \Rightarrow \left| {2\cos \theta \left( 1 \right)} \right| + \dfrac{4}{2}\cot \theta \left[ {{{\cos }^2}\left( {\dfrac{\theta }{2}} \right) + {{\sin }^2}\left( {\dfrac{\theta }{2}} \right) + 2\sin \left( {\dfrac{\theta }{2}} \right)\cos \left( {\dfrac{\theta }{2}} \right)} \right]$
Use the fourth formula which is in the hint.
$ \Rightarrow \left| {2\cos \theta } \right| + 2\cot \theta \left[ {1 + \sin \theta } \right]$
Simplify the bracket.
$ \Rightarrow \left| {2\cos \theta } \right| + 2\cot \theta + 2\cot \theta \sin \theta $
As we know, $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$.
$ \Rightarrow \left| {2\cos \theta } \right| + 2\cot \theta + 2\dfrac{{\cos \theta }}{{\sin \theta }}\sin \theta $
Here, $\sin \theta $ get cancelled.
$ \Rightarrow 2\left[ {\left| {\cos \theta } \right| + \cot \theta + \cos \theta } \right]$
In both the 2 & 3 quadrants we have given the value of cos is negative.
As the first cos is in the mod. So, there is no need to change the sign of that term.
$ \Rightarrow 2\left[ {\left| {\cos \theta } \right| + \cot \theta - \cos \theta } \right]$
$ \Rightarrow 2\left[ {\cot \theta } \right]$
$ \Rightarrow 2\cot \theta $
Therefore, option (C) is the correct answer.
Note:
- To solve this kind of problem, students should be familiar with trigonometric basic formulas, identities, reciprocal identities, angle sum & difference identities.
- Students should know about trigonometric functions and the signs of main trigonometric functions sin, cos, tan in different quadrants.
- Important thing is that students should know the values of all trigonometric functions at various degrees.
Here we can use some trigonometric formulas.
1) $\sin 2\theta = 2\sin \theta \cos \theta $
2) $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
3) $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
4) ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Complete step by step answer:
1) Let us simplify the given term.
We have,
$ = \sqrt {4{{\cos }^4}\theta + {{\sin }^2}2\theta } + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$
Let us simplify first the terms in the square root.
We can use 1) formula.
We get,
$ \Rightarrow \sqrt {4{{\cos }^4}\theta + {{\left( {2\sin \theta \cos \theta } \right)}^2}} + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$
Simplify the remaining term.
Let us use 2) formula.
Now we get,
$ \Rightarrow \sqrt {4{{\cos }^4}\theta + {{\left( {2\sin \theta \cos \theta } \right)}^2}} + 4\cot \theta {\left[ {\cos \dfrac{\pi }{4}\cos \dfrac{\theta }{2} + \sin \dfrac{\pi }{4}\sin \dfrac{\theta }{2}} \right]^2}$
Now, we can put the values of cos and sin.
As we know the values of, $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and$\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$.
Substitute these values.
$ \Rightarrow \sqrt {4{{\cos }^4}\theta + {{\left( {2\sin \theta \cos \theta } \right)}^2}} + 4\cot \theta {\left[ {\dfrac{1}{{\sqrt 2 }}\cos \dfrac{\theta }{2} + \dfrac{1}{{\sqrt 2 }}\sin \dfrac{\theta }{2}} \right]^2}$
Consider the term which is under the square root and is in the bracket.
Take the square of that term.
See, $\dfrac{1}{{\sqrt 2 }}$ is in both the terms.
So, let us write it outside the bracket as a common term.
So we get,
$ \Rightarrow \sqrt {4{{\cos }^4}\theta + \left( {4{{\sin }^2}\theta {{\cos }^2}\theta } \right)} + \dfrac{4}{{{{\left( {\sqrt 2 } \right)}^2}}}\cot \theta {\left[ {\cos \left( {\dfrac{\theta }{2}} \right) + \sin \left( {\dfrac{\theta }{2}} \right)} \right]^2}$
Consider the term which is under the square root.
Let us take common terms outside the bracket.
So we get,
$ \Rightarrow \sqrt {4{{\cos }^2}\theta \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)} + \dfrac{4}{{{{\left( {\sqrt 2 } \right)}^2}}}\cot \theta {\left[ {\cos \left( {\dfrac{\theta }{2}} \right) + \sin \left( {\dfrac{\theta }{2}} \right)} \right]^2}$
While removing the square root so, the value which is in the square root will be in the mod.
After that, we will use the formula to simplify the bracket:${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.
$ \Rightarrow \left| {2\cos \theta \left( 1 \right)} \right| + \dfrac{4}{2}\cot \theta \left[ {{{\cos }^2}\left( {\dfrac{\theta }{2}} \right) + {{\sin }^2}\left( {\dfrac{\theta }{2}} \right) + 2\sin \left( {\dfrac{\theta }{2}} \right)\cos \left( {\dfrac{\theta }{2}} \right)} \right]$
Use the fourth formula which is in the hint.
$ \Rightarrow \left| {2\cos \theta } \right| + 2\cot \theta \left[ {1 + \sin \theta } \right]$
Simplify the bracket.
$ \Rightarrow \left| {2\cos \theta } \right| + 2\cot \theta + 2\cot \theta \sin \theta $
As we know, $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$.
$ \Rightarrow \left| {2\cos \theta } \right| + 2\cot \theta + 2\dfrac{{\cos \theta }}{{\sin \theta }}\sin \theta $
Here, $\sin \theta $ get cancelled.
$ \Rightarrow 2\left[ {\left| {\cos \theta } \right| + \cot \theta + \cos \theta } \right]$
In both the 2 & 3 quadrants we have given the value of cos is negative.
As the first cos is in the mod. So, there is no need to change the sign of that term.
$ \Rightarrow 2\left[ {\left| {\cos \theta } \right| + \cot \theta - \cos \theta } \right]$
$ \Rightarrow 2\left[ {\cot \theta } \right]$
$ \Rightarrow 2\cot \theta $
Therefore, option (C) is the correct answer.
Note:
- To solve this kind of problem, students should be familiar with trigonometric basic formulas, identities, reciprocal identities, angle sum & difference identities.
- Students should know about trigonometric functions and the signs of main trigonometric functions sin, cos, tan in different quadrants.
- Important thing is that students should know the values of all trigonometric functions at various degrees.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

How is gypsum formed class 10 chemistry CBSE

If the line 3x + 4y 24 0 intersects the xaxis at t-class-10-maths-CBSE

Sugar present in DNA is A Heptose B Hexone C Tetrose class 10 biology CBSE

Trending doubts
The average rainfall in India is A 105cm B 90cm C 120cm class 10 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

Indias first jute mill was established in 1854 in A class 10 social science CBSE

Indias first jute mill was established in 1854 in A class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

