Question

cotA+cosecA=3,then find sinA=?​

Answer 1

$\textbf{Given:}$

$cosecA+cotA=3$ .......(1)

$\text{We know that,}$

$\bf\;cosec^2A-cot^2A=1$

$\implies(cosecA-cotA)(cosecA+cotA)=1$

$\implies\;cosecA-cotA=\frac{1}{3}$ .......(2)

$\text{Adding (1) and (2), we get}$

$2\;cosecA=3+\frac{1}{3}$

$2\;cosecA=\frac{10}{3}$

$\implies\;cosecA=\frac{5}{3}$

$\text{Taking reciproccals, we get}$

$\bf\;sin\;A=\frac{3}{5}$

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If 6 tan A -5 = 0 find the value of( 3 Sin A - Cos A) / (5 Cos A + 9 sin A)

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Answer 2

The value of $\sin A$  is equal to $\sin A=\dfrac{3}{5}$.

Step-by-step explanation:

Given,

$\cot A+\csc A=3$               .........(1)

To find, the value of $\sin A$ = ?​

Using the trigonometric identity,

$\csc^2 A-\cot^2 A=1$

⇒  $(\csc A+\cot A)(\csc A-\cot A)=1$  

Using equation (1), we get

$(3)(\csc A-\cot A)=1$

⇒ $\csc A-\cot A =\dfrac{1}{3}$      .........(2)

Adding equations (1) and (2), we get

$\cot A+\csc A+\csc A-\cot A=3+\dfrac{1}{3}$

⇒ $\csc A+\csc A=\dfrac{9+1}{3}$

⇒ $2\csc A=\dfrac{10}{3}$

⇒ $\csc A=\dfrac{5}{3}$

∴ $\sin A=\dfrac{3}{5}$

Thus, the value of $\sin A$  is equal to $\sin A=\dfrac{3}{5}$.