
How do simplify $ \sec 35{}^\circ \operatorname{cosec}55{}^\circ -\tan 35{}^\circ \cot 55{}^\circ $ ?
Answer
536.7k+ views
Hint:
To simplify the above expression, we will use the fundamental trigonometric formulas such as $ \sec x=\dfrac{1}{\cos x} $ , $ \operatorname{cosec}x=\dfrac{1}{\sin x} $ , $ \tan x=\dfrac{\sin x}{\cos x} $ , and $ \cot x=\dfrac{\cos x}{\sin x} $ , and we will also use the identity such as $ \sin \left( 90{}^\circ -\theta \right)=\cos \theta $ , and $ \cos \left( 90{}^\circ -\theta \right)=\sin \theta $ .
Complete step by step answer:
We can see that the above-given question is of trigonometry, so we will use trigonometric identity to solve the above question.
Since, we have to simplify $ \sec 35{}^\circ \operatorname{cosec}55{}^\circ -\tan 35{}^\circ \cot 55{}^\circ $ .
We will first change all the terms into sine and cosine forms by using the trigonometric formula such as $ \sec x=\dfrac{1}{\cos x} $ , $ \operatorname{cosec}x=\dfrac{1}{\sin x} $ , $ \tan x=\dfrac{\sin x}{\cos x} $ , and $ \cot x=\dfrac{\cos x}{\sin x} $ .
So, we can write $ \sec 35{}^\circ \operatorname{cosec}55{}^\circ -\tan 35{}^\circ \cot 55{}^\circ $ as :
$ \Rightarrow \sec 35{}^\circ \operatorname{cosec}55{}^\circ -\tan 35{}^\circ \cot 55{}^\circ =\dfrac{1}{\cos 35{}^\circ }\times \dfrac{1}{\sin 55{}^\circ }-\dfrac{\sin 35{}^\circ }{\cos 35{}^\circ }\times \dfrac{\cos 55{}^\circ }{\sin 55{}^\circ } $
We can see that $ 35{}^\circ +55{}^\circ =90{}^\circ $ , so we will try to change all of angles into same angle.
Since, we know that $ \sin \left( 90{}^\circ -\theta \right)=\cos \theta $ , $ \cos \left( 90{}^\circ -\theta \right)=\sin \theta $ , so we can write $ \sin \left( 55{}^\circ \right)=\sin \left( 90{}^\circ -35{}^\circ \right)=\cos 35{}^\circ $ , $ \cos \left( 55{}^\circ \right)=\cos \left( 90{}^\circ -35{}^\circ \right)=\sin 35{}^\circ $ .
$ \Rightarrow \dfrac{1}{\cos 35{}^\circ \sin 55{}^\circ }-\dfrac{\sin 35{}^\circ \cos 55{}^\circ }{\cos 35{}^\circ \sin 55{}^\circ }=\dfrac{1}{\cos 35{}^\circ \cos 35{}^\circ }-\dfrac{\sin 35{}^\circ \sin 35{}^\circ }{\cos 35{}^\circ \cos 35{}^\circ } $
$ \Rightarrow \dfrac{1}{\cos 35{}^\circ \sin 55{}^\circ }-\dfrac{\sin 35{}^\circ \cos 55{}^\circ }{\cos 35{}^\circ \sin 55{}^\circ }=\dfrac{1}{{{\cos }^{2}}35{}^\circ }-\dfrac{{{\sin }^{2}}35{}^\circ }{{{\cos }^{2}}35{}^\circ } $
$ =\dfrac{1-{{\sin }^{2}}35{}^\circ }{{{\cos }^{2}}35{}^\circ } $
Since, we know that $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ , So we can say that $ {{\cos }^{2}}35{}^\circ =1-{{\sin }^{2}}35{}^\circ $ .
$ =\dfrac{{{\cos }^{2}}35{}^\circ }{{{\cos }^{2}}35{}^\circ } $
= 1
Hence, $ \sec 35{}^\circ \operatorname{cosec}55{}^\circ -\tan 35{}^\circ \cot 55{}^\circ $ = 1.
This is our required solution.
Note:
Students are required to note that when we are given $ \sec \theta $ , $ \operatorname{cosec}\theta $ , $ \tan \theta $ , and $ \cot \theta $ in the trigonometric expression then we always change them into $ \sin \theta $ and $ \cos \theta $ because we can easily simplify them. In the above question we have changed all the angle into $ 35{}^\circ $ , we can also solve them by changing all the angle into $ 55{}^\circ $ in the similar ways.
We can write $ \dfrac{1}{\cos 35{}^\circ }\times \dfrac{1}{\sin 55{}^\circ }-\dfrac{\sin 35{}^\circ }{\cos 35{}^\circ }\times \dfrac{\cos 55{}^\circ }{\sin 55{}^\circ } $ also as $ \Rightarrow \dfrac{1}{\cos 35{}^\circ \sin 55{}^\circ }-\dfrac{\sin 35{}^\circ \cos 55{}^\circ }{\cos 35{}^\circ \sin 55{}^\circ }=\dfrac{1}{\cos 55{}^\circ \cos 55{}^\circ }-\dfrac{\sin 55{}^\circ \sin 55{}^\circ }{\cos 55{}^\circ \cos 55{}^\circ } $
$ =\dfrac{1}{{{\cos }^{2}}55{}^\circ }-\dfrac{{{\sin }^{2}}55{}^\circ }{{{\cos }^{2}}55{}^\circ } $
$ =\dfrac{1-{{\sin }^{2}}55{}^\circ }{{{\cos }^{2}}55{}^\circ } $
$ =1 $
Students are also required to memorize the trigonometric formulas very well otherwise they will not be able to solve trigonometric question.
To simplify the above expression, we will use the fundamental trigonometric formulas such as $ \sec x=\dfrac{1}{\cos x} $ , $ \operatorname{cosec}x=\dfrac{1}{\sin x} $ , $ \tan x=\dfrac{\sin x}{\cos x} $ , and $ \cot x=\dfrac{\cos x}{\sin x} $ , and we will also use the identity such as $ \sin \left( 90{}^\circ -\theta \right)=\cos \theta $ , and $ \cos \left( 90{}^\circ -\theta \right)=\sin \theta $ .
Complete step by step answer:
We can see that the above-given question is of trigonometry, so we will use trigonometric identity to solve the above question.
Since, we have to simplify $ \sec 35{}^\circ \operatorname{cosec}55{}^\circ -\tan 35{}^\circ \cot 55{}^\circ $ .
We will first change all the terms into sine and cosine forms by using the trigonometric formula such as $ \sec x=\dfrac{1}{\cos x} $ , $ \operatorname{cosec}x=\dfrac{1}{\sin x} $ , $ \tan x=\dfrac{\sin x}{\cos x} $ , and $ \cot x=\dfrac{\cos x}{\sin x} $ .
So, we can write $ \sec 35{}^\circ \operatorname{cosec}55{}^\circ -\tan 35{}^\circ \cot 55{}^\circ $ as :
$ \Rightarrow \sec 35{}^\circ \operatorname{cosec}55{}^\circ -\tan 35{}^\circ \cot 55{}^\circ =\dfrac{1}{\cos 35{}^\circ }\times \dfrac{1}{\sin 55{}^\circ }-\dfrac{\sin 35{}^\circ }{\cos 35{}^\circ }\times \dfrac{\cos 55{}^\circ }{\sin 55{}^\circ } $
We can see that $ 35{}^\circ +55{}^\circ =90{}^\circ $ , so we will try to change all of angles into same angle.
Since, we know that $ \sin \left( 90{}^\circ -\theta \right)=\cos \theta $ , $ \cos \left( 90{}^\circ -\theta \right)=\sin \theta $ , so we can write $ \sin \left( 55{}^\circ \right)=\sin \left( 90{}^\circ -35{}^\circ \right)=\cos 35{}^\circ $ , $ \cos \left( 55{}^\circ \right)=\cos \left( 90{}^\circ -35{}^\circ \right)=\sin 35{}^\circ $ .
$ \Rightarrow \dfrac{1}{\cos 35{}^\circ \sin 55{}^\circ }-\dfrac{\sin 35{}^\circ \cos 55{}^\circ }{\cos 35{}^\circ \sin 55{}^\circ }=\dfrac{1}{\cos 35{}^\circ \cos 35{}^\circ }-\dfrac{\sin 35{}^\circ \sin 35{}^\circ }{\cos 35{}^\circ \cos 35{}^\circ } $
$ \Rightarrow \dfrac{1}{\cos 35{}^\circ \sin 55{}^\circ }-\dfrac{\sin 35{}^\circ \cos 55{}^\circ }{\cos 35{}^\circ \sin 55{}^\circ }=\dfrac{1}{{{\cos }^{2}}35{}^\circ }-\dfrac{{{\sin }^{2}}35{}^\circ }{{{\cos }^{2}}35{}^\circ } $
$ =\dfrac{1-{{\sin }^{2}}35{}^\circ }{{{\cos }^{2}}35{}^\circ } $
Since, we know that $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ , So we can say that $ {{\cos }^{2}}35{}^\circ =1-{{\sin }^{2}}35{}^\circ $ .
$ =\dfrac{{{\cos }^{2}}35{}^\circ }{{{\cos }^{2}}35{}^\circ } $
= 1
Hence, $ \sec 35{}^\circ \operatorname{cosec}55{}^\circ -\tan 35{}^\circ \cot 55{}^\circ $ = 1.
This is our required solution.
Note:
Students are required to note that when we are given $ \sec \theta $ , $ \operatorname{cosec}\theta $ , $ \tan \theta $ , and $ \cot \theta $ in the trigonometric expression then we always change them into $ \sin \theta $ and $ \cos \theta $ because we can easily simplify them. In the above question we have changed all the angle into $ 35{}^\circ $ , we can also solve them by changing all the angle into $ 55{}^\circ $ in the similar ways.
We can write $ \dfrac{1}{\cos 35{}^\circ }\times \dfrac{1}{\sin 55{}^\circ }-\dfrac{\sin 35{}^\circ }{\cos 35{}^\circ }\times \dfrac{\cos 55{}^\circ }{\sin 55{}^\circ } $ also as $ \Rightarrow \dfrac{1}{\cos 35{}^\circ \sin 55{}^\circ }-\dfrac{\sin 35{}^\circ \cos 55{}^\circ }{\cos 35{}^\circ \sin 55{}^\circ }=\dfrac{1}{\cos 55{}^\circ \cos 55{}^\circ }-\dfrac{\sin 55{}^\circ \sin 55{}^\circ }{\cos 55{}^\circ \cos 55{}^\circ } $
$ =\dfrac{1}{{{\cos }^{2}}55{}^\circ }-\dfrac{{{\sin }^{2}}55{}^\circ }{{{\cos }^{2}}55{}^\circ } $
$ =\dfrac{1-{{\sin }^{2}}55{}^\circ }{{{\cos }^{2}}55{}^\circ } $
$ =1 $
Students are also required to memorize the trigonometric formulas very well otherwise they will not be able to solve trigonometric question.
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