Answer
414.6k+ views
Hint: First we will convert $\left( {\dfrac{\pi }{2} + \theta } \right)$ into $\theta $ after that we will convert all the trigonometric ratio by using $\sec \theta = \dfrac{1}{{\cos \theta }}$,$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ after simplifying that equate the linear equation with 0 and find the value of $x$.
Complete step-by-step answer:
1:- Given $coesc\left( {\dfrac{\pi }{2} + \theta } \right) + x\cos \theta \cot \left( {\dfrac{\pi }{2} + \theta } \right) = \sin \left( {\dfrac{\pi }{2} + \theta } \right)$
Convert $\left( {\dfrac{\pi }{2} + \theta } \right)$into $\theta $ we get
$ \Rightarrow \sec \theta - x\cos \theta \tan \theta = \cos \theta $
We know that$\sec \theta = \dfrac{1}{{\cos \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ substituting these values, we get,
$ \Rightarrow \dfrac{1}{{\cos \theta }} - x\sin \theta = \cos \theta $
Multiplying $\cos \theta $ on both side we get,
$ \Rightarrow 1 - x\cos \theta \sin \theta = {\cos ^2}\theta $
We know that $1 - {\cos ^2}\theta = {\sin ^2}\theta $
$ \Rightarrow {\sin ^2}\theta - x\sin \theta \cos \theta = 0$
Taking $\sin \theta $ common
$ \Rightarrow \sin \theta \left( {\sin \theta - x\cos \theta } \right) = 0$
Now,
$
\Rightarrow \sin \theta - x\cos \theta = 0 \\
\therefore x = \tan \theta \\
$
2: - Given $x\cot \left( {\dfrac{\pi }{2} + \theta } \right) + \sin \theta \tan \left( {\dfrac{\pi }{2} + \theta } \right) + \cos ec\left( {\dfrac{\pi }{2} + \theta } \right) = 0$
Convert $\left( {\dfrac{\pi }{2} + \theta } \right)$ into $\theta $ we get
$ \Rightarrow - x\tan \theta - sin\theta \cot \theta + \sec \theta = 0$
We know that $\sec \theta = \dfrac{1}{{\cos \theta }}$,$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ substituting these values, we get,
$ \Rightarrow - x\dfrac{{\sin \theta }}{{\cos \theta }} - \cos \theta + \dfrac{1}{{\cos \theta }} = 0$
Multiplying $\cos \theta $on both side we get,
$ \Rightarrow 1 - x\sin \theta = {\cos ^2}\theta $
We know that $1 - {\cos ^2}\theta = {\sin ^2}\theta $
$ \Rightarrow {\sin ^2}\theta - x\sin \theta \cos \theta = 0$
Taking $\sin \theta $ common
$ \Rightarrow \sin \theta \left( {\sin \theta - x\cos \theta } \right) = 0$
Now,
$
\Rightarrow \sin \theta - x\cos \theta = 0 \\
\therefore x = \tan \theta \\
$
Note: Trigonometric formulas used in this question $coesc\left( {\dfrac{\pi }{2} + \theta } \right) = \sec \theta $,$\cot \left( {\dfrac{\pi }{2} + \theta } \right) = - \tan \theta $,$\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta $and must be taken into consideration. We can also say as $\sin \theta = 0$then $\theta = n\pi $. But in this question, we cannot consider $\theta = n\pi $because when we consider it $x = 0$which contradicts the question.
Complete step-by-step answer:
1:- Given $coesc\left( {\dfrac{\pi }{2} + \theta } \right) + x\cos \theta \cot \left( {\dfrac{\pi }{2} + \theta } \right) = \sin \left( {\dfrac{\pi }{2} + \theta } \right)$
Convert $\left( {\dfrac{\pi }{2} + \theta } \right)$into $\theta $ we get
$ \Rightarrow \sec \theta - x\cos \theta \tan \theta = \cos \theta $
We know that$\sec \theta = \dfrac{1}{{\cos \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ substituting these values, we get,
$ \Rightarrow \dfrac{1}{{\cos \theta }} - x\sin \theta = \cos \theta $
Multiplying $\cos \theta $ on both side we get,
$ \Rightarrow 1 - x\cos \theta \sin \theta = {\cos ^2}\theta $
We know that $1 - {\cos ^2}\theta = {\sin ^2}\theta $
$ \Rightarrow {\sin ^2}\theta - x\sin \theta \cos \theta = 0$
Taking $\sin \theta $ common
$ \Rightarrow \sin \theta \left( {\sin \theta - x\cos \theta } \right) = 0$
Now,
$
\Rightarrow \sin \theta - x\cos \theta = 0 \\
\therefore x = \tan \theta \\
$
2: - Given $x\cot \left( {\dfrac{\pi }{2} + \theta } \right) + \sin \theta \tan \left( {\dfrac{\pi }{2} + \theta } \right) + \cos ec\left( {\dfrac{\pi }{2} + \theta } \right) = 0$
Convert $\left( {\dfrac{\pi }{2} + \theta } \right)$ into $\theta $ we get
$ \Rightarrow - x\tan \theta - sin\theta \cot \theta + \sec \theta = 0$
We know that $\sec \theta = \dfrac{1}{{\cos \theta }}$,$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ substituting these values, we get,
$ \Rightarrow - x\dfrac{{\sin \theta }}{{\cos \theta }} - \cos \theta + \dfrac{1}{{\cos \theta }} = 0$
Multiplying $\cos \theta $on both side we get,
$ \Rightarrow 1 - x\sin \theta = {\cos ^2}\theta $
We know that $1 - {\cos ^2}\theta = {\sin ^2}\theta $
$ \Rightarrow {\sin ^2}\theta - x\sin \theta \cos \theta = 0$
Taking $\sin \theta $ common
$ \Rightarrow \sin \theta \left( {\sin \theta - x\cos \theta } \right) = 0$
Now,
$
\Rightarrow \sin \theta - x\cos \theta = 0 \\
\therefore x = \tan \theta \\
$
Note: Trigonometric formulas used in this question $coesc\left( {\dfrac{\pi }{2} + \theta } \right) = \sec \theta $,$\cot \left( {\dfrac{\pi }{2} + \theta } \right) = - \tan \theta $,$\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta $and must be taken into consideration. We can also say as $\sin \theta = 0$then $\theta = n\pi $. But in this question, we cannot consider $\theta = n\pi $because when we consider it $x = 0$which contradicts the question.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)