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Question: How do you prove \(\tan \left( x \right)\sin \left( x \right)+\cos \left( x \right)=\sec \left( x \r...

How do you prove tan⁡(x)sin⁡(x)+cos⁡(x)=sec⁡(x)\tan \left( x \right)\sin \left( x \right)+\cos \left( x \right)=\sec \left( x \right).

Explanation

Solution

The above given question is of trigonometric identities. So, we will use the fundamental trigonometric formulas such as tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}, sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x} and we will also use the identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1, to prove the above expression.

Complete step-by-step solution:
We can see that above given question is of trigonometric identity and so we will use trigonometric formulas to prove the above result tan⁡(x)sin⁡(x)+cos⁡(x)=sec⁡(x)\tan \left( x \right)\sin \left( x \right)+\cos \left( x \right)=\sec \left( x \right).
Since, we have to prove that tan⁡(x)sin⁡(x)+cos⁡(x)=sec⁡(x)\tan \left( x \right)\sin \left( x \right)+\cos \left( x \right)=\sec \left( x \right).
We will make the LHS term equal to the RHS.
From LHS of the equation we know that LHS = tan⁡(x)sin⁡(x)+cos⁡(x)\tan \left( x \right)\sin \left( x \right)+\cos \left( x \right)
Since, we know that tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} . So, we will put sin⁡xcos⁡x\dfrac{\sin x}{\cos x} in place of tan⁡x\tan x.
⇒LHS=tan⁡(x)sin⁡(x)+cos⁡(x)=sin⁡xcos⁡x×sin⁡x+cos⁡x\Rightarrow LHS=\tan \left( x \right)\sin \left( x \right)+\cos \left( x \right)=\dfrac{\sin x}{\cos x}\times \sin x+\cos x
Now, we will take cos x as LCM then we will get:
⇒tan⁡(x)sin⁡(x)+cos⁡(x)=sin⁡x×sin⁡x+cos⁡x×cos⁡xcos⁡x\Rightarrow \tan \left( x \right)\sin \left( x \right)+\cos \left( x \right)=\dfrac{\sin x\times \sin x+\cos x\times \cos x}{\cos x}
⇒tan⁡(x)sin⁡(x)+cos⁡(x)=sin⁡2x+cos⁡2xcos⁡x\Rightarrow \tan \left( x \right)\sin \left( x \right)+\cos \left( x \right)=\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos x}
Now, we will use the trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1,
⇒tan⁡(x)sin⁡(x)+cos⁡(x)=1cos⁡x\Rightarrow \tan \left( x \right)\sin \left( x \right)+\cos \left( x \right)=\dfrac{1}{\cos x}
Now, we know that sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x}, so we will put sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x}.
⇒tan⁡(x)sin⁡(x)+cos⁡(x)=sec⁡x\Rightarrow \tan \left( x \right)\sin \left( x \right)+\cos \left( x \right)=\sec x
Since, LHS = sec x which is equal to RHS.
So, LHS = RHS
Hence, proved.
This is our required solution.

Note: Students are required to note that when we are given sec⁡θ\sec \theta , cosec⁡θ\operatorname{cosec}\theta , tan⁡θ\tan \theta , and cot⁡θ\cot \theta in the trigonometric expression then we always change them into sin⁡θ\sin \theta and cos⁡θ\cos \theta . Also, students are required to memorize all the trigonometric formulas otherwise they will not be able to prove the above question.