Question

If $4\sin A-3\cos A=0$, find $\sin A,\cos A,\sec A$ and $\text{cosec A}$.

Answer 1

Let $△PQR$ be a right angled triangle where $\angle Q={90}^0$ and $\angle R=A$ as shown in the above figure:

Now it is given that $4\sin A-3\cos A=0$ that is

$4\sin A=3\cos A\Rightarrow \frac{\sin A}{\cos A}=\frac{3}{4}\Rightarrow \tan A=\frac{3}{4}$

We know that, in a right angled triangle, $\tan \theta$ is equal to opposite side over adjacent side that is $\tan \theta =\frac{Oppositeside}{Adjacentside}$, therefore, opposite side $PQ=3$ and adjacent side $QR=4$.

Now, using pythagoras theorem in $△PQR$, we have

$P{R}^2=P{Q}^2+Q{R}^2=3^2+4^2=9+16=25\Rightarrow PR=\sqrt{25}=5$

Therefore, the hypotenuse $PR=5$.

We know that, in a right angled triangle,

$\sin \theta$ is equal to opposite side over hypotenuse that is $\sin \theta =\frac{Oppositeside}{Hypotenuse}$ and

$\cos \theta$ is equal to adjacent side over hypotenuse that is $\cos \theta =\frac{Adjacentside}{Hypotenuse}$

Here, we have opposite side $PQ=3$, adjacent side $QR=4$ and the hypotenuse $PR=5$, therefore, the trignometric ratios of angle $A$ can be determined as follows:

$\sin A=\frac{Oppositeside}{Hypotenuse}=\frac{PQ}{PR}=\frac{3}{5}$

$\cos A=\frac{Adjacentside}{Hypotenuse}=\frac{QR}{PR}=\frac{4}{5}$

$cosecA=\frac{1}{\sin A}=\frac{1}{\frac{3}{5}}=1\times \frac{5}{3}=\frac{5}{3}$

$\sec A=\frac{1}{\cos A}=\frac{1}{\frac{4}{5}}=1\times \frac{5}{4}=\frac{5}{4}$

Hence, $\sin A=\frac{3}{5}$, $\cos A=\frac{4}{5}$, $\csc A=\frac{5}{3}$ and $\sec A=\frac{5}{4}$.