prove the following identity : 1.sin alpha cos (90°- alpha) +cos alpha sin (90°- alpha) = 1 2. tan² Alpha + 1 = sec² alpha ; ctg² alpha + 1 = cosec 3. tan (90°

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\text{No. 1}\\
\sin{\alpha}\cos{(90\degree-\alpha)}+\cos{\alpha}\sin{(90\degree-\alpha)}=1\\
\text{Proof:}\\
\text{We know that}\\
\sin{(90\degree-\alpha)}=\cos{\alpha}\\
\cos{(90\degree-\alpha)}=\sin{\alpha}\\
\text{also, we have a trigonometric identity}\\
\sin^2{\alpha}+\cos^2{\alpha}=1\\
\text{So that, we conclude that}\\
\sin{\alpha}\cos{(90\degree-\alpha)}+\cos{\alpha}\sin{(90\degree-\alpha)}\\
=\sin{\alpha}\sin{\alpha}+\cos{\alpha}\cos{\alpha}\\
=\sin^2{\alpha}+\cos^2{\alpha}\\
=1\\
\text{Q.E.D.}\\
\\
\text{No. 2}\\
\text{The proof is actually quiet easy}\\
\text{From trigonometric identity, we know that}\\
\sin^2{\alpha}+\cos^2{\alpha}=1\\
\text{Devide both sides with }\cos^2{\alpha}\text{ then we have}\\
\frac{\sin^2{\alpha}}{\cos^2{\alpha}}+\frac{\cos^2{\alpha}}{\cos^2{\alpha}}=\frac{1}{\cos^2{\alpha}}\\
\Leftrightarrow \tan^2{\alpha}+1=\sec^2{\alpha}\\
\text{Q.E.D}\\
\\
\text{Analogue with that, you can also prove that}\\
\cot^2{\alpha}+1=\csc^2{\alpha}\\
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