Class 9: Complementary Angles

$\displaystyle \textbf{Complementary Angles}$
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$\displaystyle \textbf{Introduction}$
$\displaystyle \text{We have studied the trigonometrical ratios of standard angles }0^\circ,30^\circ,45^\circ,60^\circ\text{ and }90^\circ.$
$\displaystyle \text{We shall now study the trigonometrical ratios of complementary angles and their applications.}$
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$\displaystyle \textbf{Concept of Complementary Angles}$
$\displaystyle \text{Two acute angles are complementary if their sum is }90^\circ.$
$\displaystyle 30^\circ+60^\circ=90^\circ$
$\displaystyle \therefore 30^\circ\text{ and }60^\circ\text{ are complementary angles.}$
$\displaystyle 70^\circ+20^\circ=90^\circ$
$\displaystyle \therefore 70^\circ\text{ and }20^\circ\text{ are complementary angles.}$
$\displaystyle x^\circ+(90-x)^\circ=90^\circ$
$\displaystyle \therefore x^\circ\text{ and }(90-x)^\circ\text{ are complementary angles.}$
$\displaystyle \text{The complement of }48^\circ=90^\circ-48^\circ=42^\circ$
$\displaystyle \text{The complement of }\theta^\circ=(90-\theta)^\circ$
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$\displaystyle \textbf{Complementary Angles for Sine and Cosine}$
$\displaystyle \text{Consider a right-angled triangle }ABC\text{ in which }\angle B=90^\circ.$
$\displaystyle \text{Let }\angle ACB=\theta.$
$\displaystyle \therefore \angle CAB=90^\circ-\theta$
$\displaystyle \sin\theta=\sin\angle ACB=\frac{AB}{AC}$
$\displaystyle \cos(90^\circ-\theta)=\cos\angle CAB=\frac{AB}{AC}$
$\displaystyle \therefore \boxed{\cos(90^\circ-\theta)=\sin\theta}$
$\displaystyle \cos\theta=\cos\angle ACB=\frac{BC}{AC}$
$\displaystyle \sin(90^\circ-\theta)=\sin\angle CAB=\frac{BC}{AC}$
$\displaystyle \therefore \boxed{\sin(90^\circ-\theta)=\cos\theta}$
$\displaystyle \textbf{Hence,}$
$\displaystyle \cos(90^\circ-\theta)=\sin\theta$
$\displaystyle \sin(90^\circ-\theta)=\cos\theta$

$\displaystyle \textbf{Complementary Angles for Tangent and Cotangent}$
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$\displaystyle \text{Consider a right-angled triangle }ABC\text{ in which }\angle B=90^\circ.$
$\displaystyle \text{Let }\angle ACB=\theta,\quad \therefore \angle CAB=90^\circ-\theta.$
$\displaystyle \tan\theta=\tan\angle ACB=\frac{AB}{BC}$
$\displaystyle \cot(90^\circ-\theta)=\cot\angle CAB=\frac{AB}{BC}$
$\displaystyle \therefore \cot(90^\circ-\theta)=\tan\theta$
$\displaystyle \text{Also, }\cot\theta=\cot\angle ACB=\frac{BC}{AB}$
$\displaystyle \tan(90^\circ-\theta)=\tan\angle CAB=\frac{BC}{AB}$
$\displaystyle \therefore \tan(90^\circ-\theta)=\cot\theta$
$\displaystyle \textbf{Thus,}$
$\displaystyle \cot(90^\circ-\theta)=\tan\theta$
$\displaystyle \tan(90^\circ-\theta)=\cot\theta$
$\displaystyle \textbf{Examples:}$
$\displaystyle \cot(90^\circ-20^\circ)=\tan20^\circ$
$\displaystyle \cot(90^\circ-57^\circ)=\tan57^\circ$
$\displaystyle \tan(90^\circ-70^\circ)=\cot70^\circ$
$\displaystyle \tan(90^\circ-25^\circ)=\cot25^\circ$
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$\displaystyle \textbf{Complementary Angles for Secant and Cosecant}$
$\displaystyle \sec\theta=\sec\angle ACB=\frac{AC}{BC}$
$\displaystyle \mathrm{cosec}(90^\circ-\theta)=\mathrm{cosec}\angle CAB=\frac{AC}{BC}$
$\displaystyle \therefore \mathrm{cosec}(90^\circ-\theta)=\sec\theta$
$\displaystyle \text{Also, }\mathrm{cosec}\theta=\mathrm{cosec}\angle ACB=\frac{AC}{AB}$
$\displaystyle \sec(90^\circ-\theta)=\sec\angle CAB=\frac{AC}{AB}$
$\displaystyle \therefore \sec(90^\circ-\theta)=\mathrm{cosec}\theta$
$\displaystyle \textbf{Thus,}$
$\displaystyle \mathrm{cosec}(90^\circ-\theta)=\sec\theta$
$\displaystyle \sec(90^\circ-\theta)=\mathrm{cosec}\theta$
$\displaystyle \textbf{Examples:}$
$\displaystyle \mathrm{cosec}(90^\circ-40^\circ)=\sec40^\circ$
$\displaystyle \mathrm{cosec}(90^\circ-67^\circ)=\sec67^\circ$
$\displaystyle \sec(90^\circ-35^\circ)=\mathrm{cosec}35^\circ$
$\displaystyle \sec(90^\circ-82^\circ)=\mathrm{cosec}82^\circ$