# If we have a trigonometric equation $\dfrac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\dfrac{1}{4}$ , then find the value of $\sin \theta $ .

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**Hint:**In question it is asked that, we have to find the value of $\sin \theta $ given that $\dfrac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\dfrac{1}{4}$.

So, to do so we will use identities and properties of trigonometric ratios such as $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$ so as to obtain the $\sin \theta $.

**Complete step-by-step solution:**We know that \[sin\text{ }\theta \text{ },\text{ }cos\text{ }\theta \text{ },\text{ }tan\text{ }\theta \text{ },\text{ }cot\text{ }\theta \text{ },\text{ }sec\text{ }\theta \text{ }and\text{ }cosec\text{ }\theta \] are trigonometric function, where \[\theta \] is the angle made by the hypotenuse with the base of triangle.

Now, in question it is given that $\dfrac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\dfrac{1}{4}$.

Now, also we know that tan A equal to the ratio of the sine function and cos A function that is $\tan A=\dfrac{\sin A}{\cos A}$.

And, also $sec\theta $ is equals to reciprocal of trigonometric function $\cos \theta $ that is $\sec \theta =\dfrac{1}{\cos \theta }$.

so, we can write $\dfrac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\dfrac{1}{4}$ as,

$\dfrac{\dfrac{1}{\cos \theta }-\dfrac{\sin A}{\cos A}}{\dfrac{1}{\cos \theta }+\dfrac{\sin A}{\cos A}}=\dfrac{1}{4}$.

On taking L.C.M in numerator an denominator, we get

$\dfrac{\dfrac{1-\sin \theta }{\cos \theta }}{\dfrac{1+\sin \theta }{\cos \theta }}=\dfrac{1}{4}$

Taking 4 from the denominator of right hand side to numerator of left hand side, $\dfrac{1+\sin \theta }{\cos \theta }$ from numerator of right hand side of left hand side to denominator of right hand side using cross multiplication, we get

$4\cdot \left( \dfrac{1-\sin \theta }{\cos \theta } \right)=\dfrac{1+\sin \theta }{\cos \theta }$……………........( i )

On solving brackets,

$\dfrac{4-4\sin \theta }{\cos \theta }=\dfrac{1+\sin \theta }{\cos \theta }$

On simplifying equation, we get

$4-4\sin \theta =1+\sin \theta $

Taking, $4\sin \theta $ from left hand side to right hand side and 1 from right hand side to left hand side,

$4-1=\sin \theta +4\sin \theta $

On solving, we get

$3=5\sin \theta $

Taking, 4 from the numerator of right hand side to denominator of left hand side, we get

$\sin \theta =\dfrac{3}{5}$

**Hence, if $\dfrac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\dfrac{1}{4}$ , then the value of $\sin \theta $ is equals to $\dfrac{3}{5}$.**

**Note:**One must know all trigonometric identities, properties, and relationships between trigonometric functions. While solving the question always use the most appropriate substitution of trigonometric relation which directly leads to the result. There may be calculation mistakes in cross multiplication, so be careful while solving an expression.