Question

How do simplify $ \sec 35{}^\circ \operatorname{cosec}55{}^\circ -\tan 35{}^\circ \cot 55{}^\circ $ ?

Answer 1

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.