Question.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
We have $\operatorname{cosec}^{2} A-\cot ^{2} A=1$
$\Rightarrow \operatorname{cosec}^{2} A=1+\cot ^{2} A$
$\Rightarrow(\operatorname{cosec} A)^{2}=\cot ^{2} A+1$
$\Rightarrow\left(\frac{1}{\sin A}\right)^{2}=\cot ^{2} A+1$
$\Rightarrow(\sin \mathrm{A})^{2}=\frac{\mathbf{1}}{\cot ^{2} \mathbf{A}+\mathbf{1}}$
$\Rightarrow \sin A=\pm \frac{1}{\sqrt{\cot ^{2} A+1}}$
We reject negative value of sin A for acute angle
A. Therefore, $\sin \mathrm{A}=\frac{\mathbf{1}}{\sqrt{\cot ^{2} \mathbf{A}+\mathbf{1}}} \tan \mathrm{A}=\frac{\mathbf{1}}{\cot \mathbf{A}}$
We have $\sec ^{2} A-\tan ^{2} A=1$
$\Rightarrow \sec ^{2} A=1+\tan ^{2} A$
$=1+\frac{1}{\cot ^{2} \mathbf{A}}=\frac{\cot ^{2} \mathbf{A}+1}{\cot ^{2} \mathbf{A}}$
$\Rightarrow \sec A=\frac{\sqrt{\cot ^{2} A+1}}{\cot A}$
We have $\operatorname{cosec}^{2} A-\cot ^{2} A=1$
$\Rightarrow \operatorname{cosec}^{2} A=1+\cot ^{2} A$
$\Rightarrow(\operatorname{cosec} A)^{2}=\cot ^{2} A+1$
$\Rightarrow\left(\frac{1}{\sin A}\right)^{2}=\cot ^{2} A+1$
$\Rightarrow(\sin \mathrm{A})^{2}=\frac{\mathbf{1}}{\cot ^{2} \mathbf{A}+\mathbf{1}}$
$\Rightarrow \sin A=\pm \frac{1}{\sqrt{\cot ^{2} A+1}}$
We reject negative value of sin A for acute angle
A. Therefore, $\sin \mathrm{A}=\frac{\mathbf{1}}{\sqrt{\cot ^{2} \mathbf{A}+\mathbf{1}}} \tan \mathrm{A}=\frac{\mathbf{1}}{\cot \mathbf{A}}$
We have $\sec ^{2} A-\tan ^{2} A=1$
$\Rightarrow \sec ^{2} A=1+\tan ^{2} A$
$=1+\frac{1}{\cot ^{2} \mathbf{A}}=\frac{\cot ^{2} \mathbf{A}+1}{\cot ^{2} \mathbf{A}}$
$\Rightarrow \sec A=\frac{\sqrt{\cot ^{2} A+1}}{\cot A}$
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