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Question: Prove that \(\left( \sin \alpha +\cos \alpha \right)\left( \tan \alpha +\cot \alpha \right)=\sec \al...

Prove that (sin⁡α+cos⁡α)(tan⁡α+cot⁡α)=sec⁡α+csc⁡α\left( \sin \alpha +\cos \alpha \right)\left( \tan \alpha +\cot \alpha \right)=\sec \alpha +\csc \alpha .

Explanation

Solution

We have a sum of two terms on the left-hand side of (tan⁡α+cot⁡α)\left( \tan \alpha +\cot \alpha \right). We simplify tan⁡α=sin⁡αcos⁡α\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha } and cot⁡α=cos⁡αsin⁡α\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha } on both numerator and denominator. Then we add them as the LCM of the denominators are the multiplication of them. Then we use the identity of (sin⁡2α+cos⁡2α)=1\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \right)=1 to find the solution of the numerator. We also need to mention the conditions.

Complete step-by-step solution:
We simplify the addition part of (tan⁡α+cot⁡α)\left( \tan \alpha +\cot \alpha \right) by breaking it as tan⁡α=sin⁡αcos⁡α\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha } and cot⁡α=cos⁡αsin⁡α\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha }. We replace the values as and get
(tan⁡α+cot⁡α) =sin⁡αcos⁡α+cos⁡αsin⁡α =sin⁡2α+cos⁡2αsin⁡αcos⁡α \begin{aligned} & \left( \tan \alpha +\cot \alpha \right) \\\ & =\dfrac{\sin \alpha }{\cos \alpha }+\dfrac{\cos \alpha }{\sin \alpha } \\\ & =\dfrac{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }{\sin \alpha \cos \alpha } \\\ \end{aligned}
We replace the value with (sin⁡2α+cos⁡2α)=1\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \right)=1.
Therefore, (tan⁡α+cot⁡α)=sin⁡2α+cos⁡2αsin⁡αcos⁡α=1sin⁡αcos⁡α\left( \tan \alpha +\cot \alpha \right)=\dfrac{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }{\sin \alpha \cos \alpha }=\dfrac{1}{\sin \alpha \cos \alpha }
We multiply 1sin⁡αcos⁡α\dfrac{1}{\sin \alpha \cos \alpha } with (sin⁡α+cos⁡α)\left( \sin \alpha +\cos \alpha \right).
So, (sin⁡α+cos⁡α)×1sin⁡αcos⁡α=sin⁡α+cos⁡αsin⁡αcos⁡α\left( \sin \alpha +\cos \alpha \right)\times \dfrac{1}{\sin \alpha \cos \alpha }=\dfrac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }.
We break the individual terms and get sin⁡α+cos⁡αsin⁡αcos⁡α=sin⁡αsin⁡αcos⁡α+cos⁡αsin⁡αcos⁡α\dfrac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }=\dfrac{\sin \alpha }{\sin \alpha \cos \alpha }+\dfrac{\cos \alpha }{\sin \alpha \cos \alpha }.
Bu eliminating common terms we get sin⁡α+cos⁡αsin⁡αcos⁡α=1cos⁡α+1sin⁡α\dfrac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }=\dfrac{1}{\cos \alpha }+\dfrac{1}{\sin \alpha }
We know that the relations for inverse are 1cos⁡α=sec⁡α,1sin⁡α=csc⁡α\dfrac{1}{\cos \alpha }=\sec \alpha ,\dfrac{1}{\sin \alpha }=\csc \alpha
The equation becomes sin⁡α+cos⁡αsin⁡αcos⁡α=1cos⁡α+1sin⁡α=sec⁡α+csc⁡α\dfrac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }=\dfrac{1}{\cos \alpha }+\dfrac{1}{\sin \alpha }=\sec \alpha +\csc \alpha .
Thus proved (sin⁡α+cos⁡α)(tan⁡α+cot⁡α)=sec⁡α+csc⁡α\left( \sin \alpha +\cos \alpha \right)\left( \tan \alpha +\cot \alpha \right)=\sec \alpha +\csc \alpha .

Note: It is important to remember the condition to eliminate the cos⁡α,sin⁡α\cos \alpha ,\sin \alpha from both denominator and numerator. No domain is given for the variable xx. The value of tan⁡x≠0\tan x\ne 0 is essential. The simplified condition will be x≠nπ,n∈Zx\ne n\pi ,n\in \mathbb{Z}.