Let's first prove the identity cos θ= cot θcsc θ. We will use basic identities to show how the right-hand side is equal to the left-hand side.
\cos \theta \stackrel{?}= \dfrac{\cot \theta}{\csc \theta}
\cos \theta \stackrel{?}= \cot \theta\left(\dfrac{1}{\csc \theta}\right)
cot(θ) = cos(θ)/sin(θ)
\cos \theta \stackrel{?}= {\color{#0000FF}{\dfrac{\cos \theta}{\sin \theta}}}\left(\dfrac{1}{\csc \theta}\right)
\cos \theta \stackrel{?}= \dfrac{\cos \theta}{\sin \theta}\left(\dfrac{1}{{\color{#0000FF}{\frac{1}{\sin \theta}}}}\right)
\cos \theta \stackrel{?}= \dfrac{\cos \theta}{\sin \theta}\left(\dfrac{\sin \theta}{1}\right)
\cos \theta \stackrel{?}= \dfrac{\cos \theta \cdot \sin \theta}{\sin \theta}
\cos \theta \stackrel{?}= \dfrac{\cos \theta \cdot \cancel{{\color{#FF0000}{\sin \theta}}}}{\cancel{{\color{#FF0000}{\sin \theta}}}}
cos θ = cos θ ✓
Having proved the identity cos θ = cot θcsc θ, we can prove the identity sec θ= csc θcot θ. This time, we will start on the left-hand side and arrive to the right-hand side.
\sec \theta \stackrel{?}=\dfrac{\csc \theta}{\cot \theta}
{\color{#FF0000}{\dfrac{1}{\cos \theta}}} \stackrel{?}=\dfrac{\csc \theta}{\cot \theta}
\dfrac{1}{{\color{#0000FF}{\frac{\cot \theta}{\csc \theta}}}} \stackrel{?}=\dfrac{\csc \theta}{\cot \theta}
csc θ/cot θ = csc θ/cot θ ✓
By following similar procedures, we can prove that tan θ = 1cot θ.
\tan \theta \stackrel{?}= \dfrac{1}{\cot \theta}
\tan \theta \stackrel{?}= \dfrac{1}{{\color{#0000FF}{ \frac{\cos \theta}{\sin \theta} }}}
\tan \theta \stackrel{?}= \dfrac{\sin \theta}{\cos \theta}
tan θ = tan θ ✓
Now we are ready to move to simplifying the above fractions. Let's start with cos θ.
cos θ = 43/- 53
cos θ = - 43/53
cos θ = 4/5
Let's now follow a similar procedure for sec θ.
sec θ = - 53/43
sec θ =- 53/43
sec θ = 5/4
