[NCERT] Class 10 Ex 8.4 - Q1-Express the trigonometric ratios sinA, secA and tanA in term of cotA

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Exercise 8.1

Question1: Express the trigonometric ratios sinA, secA and tanA in terms of cotA.
Solution:  In the case of sinA, we have to use the trigonometries identities formula which has cot function

$i.e., cose{c}^{2}A - co{t}^{2}A = 1$
$cose{c}^{2}A = 1 + co{t}^{2}A$
$\frac{1}{{\sin }^{2}A} = 1 + co{t}^{2}A$
$\frac{1}{ 1 + co{t}^{2}A} = {\sin }^{2}A ........... [Eqn-1]$
$Therefore, sinA = \pm \frac{1}{\sqrt{1 + co{t}^{2}A}}$
$In case of secA, we have to use equation 1 below$
$\frac{1}{ 1 + co{t}^{2}A} = {\sin }^{2}A = 1-co{s}^{2}A [\because {\sin }^{2}A + co{s}^{2}A = 1]$
$\Rightarrow co{s}^{2}A = 1 - \frac{1}{ 1 + co{t}^{2}A} = \frac{1 + co{t}^{2}A - 1}{ 1 + co{t}^{2}A} = \frac{co{t}^{2}A}{ 1 + co{t}^{2}A}$
$\frac{1}{{\sec }^{2}A} = \frac{co{t}^{2}A}{ 1 + co{t}^{2}A}$

In case of tanA, as tanA is the reciprocal of cotA

Therefore, tanA =$\frac{1}{\cot A}$

1. Class 10 Ex-8.1 Q2 Solution

2. Class 10 Ex-8.1 Q3 Solution

3. Class 10 Ex-8.1 Q4 Solution

4. Class 10 Ex-8.1 Q5 Solution