Buktikan bahwa : a. [tex] \tan \alpha + \cot \alpha = \sec \alpha \csc \alpha [/tex] b. [tex] \frac{ \sin\alpha }{ \cos\alpha } + \frac{ \cos\alpha }{ \

Soal bagian a.
Tips: Gunakan definisi fungsi trigonometri bahwa:
[tex]\boxed{\tan x=\frac{\sin x}{\cos x}\text{ dan }\cot x=\frac{\cos x}{\sin x}}[/tex]
Hasil berikut memberikan:
[tex]$\begin{align}\tan\alpha+\cot\alpha&=\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha} \\ &=\frac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha\cos\alpha} \\ &=\frac{1}{\sin\alpha\cos\alpha}&&\text{Identitas: }sin^2\alpha+\cos^2\alpha=1 \\ &=\frac{1}{\sin\alpha}\times\frac{1}{\cos\alpha} \\ &=\csc \alpha\sec\alpha\end{align}[/tex]

Penjabaran yang sama dapat digunakan untuk b, yang diperoleh:
[tex]$\begin{align}\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha} &=\frac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha\cos\alpha} \\ &=\frac{1}{\sin\alpha\cos\alpha} \\ &=\frac{1}{\sin\alpha}\times\frac{1}{\cos\alpha} \\ &=\csc \alpha\sec\alpha \\ &=\sqrt{\csc^2\alpha}\times\sqrt{\sec^2\alpha} \\ &=\sqrt{1+\cot^2\alpha}\times\sqrt{1+\tan^2\alpha} \\ &=\sqrt{(1+\tan^2\alpha)(1+\cot^2\alpha)}\end{align}[/tex]
Berdasarkan identitas trigonometri:
1 + tan² α = sec² α
1 + cot² α = csc² α