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Question: If we have a trigonometric equation \(\dfrac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\...

If we have a trigonometric equation sec⁡θ−tan⁡θsec⁡θ+tan⁡θ=14\dfrac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\dfrac{1}{4} , then find the value of sin⁡θ\sin \theta .

Explanation

Solution

In question it is asked that, we have to find the value of sin⁡θ\sin \theta given that sec⁡θ−tan⁡θsec⁡θ+tan⁡θ=14\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⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } so as to obtain the sin⁡θ\sin \theta .

Complete step-by-step solution:
We know that sin θ , cos θ , tan θ , cot θ , sec θ and cosec θsin\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 θ\theta is the angle made by the hypotenuse with the base of triangle.
Now, in question it is given that sec⁡θ−tan⁡θsec⁡θ+tan⁡θ=14\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=sin⁡Acos⁡A\tan A=\dfrac{\sin A}{\cos A}.
And, also secθsec\theta is equals to reciprocal of trigonometric function cos⁡θ\cos \theta that is sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta }.
so, we can write sec⁡θ−tan⁡θsec⁡θ+tan⁡θ=14\dfrac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\dfrac{1}{4} as,
1cos⁡θ−sin⁡Acos⁡A1cos⁡θ+sin⁡Acos⁡A=14\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
1−sin⁡θcos⁡θ1+sin⁡θcos⁡θ=14\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, 1+sin⁡θcos⁡θ\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⋅(1−sin⁡θcos⁡θ)=1+sin⁡θcos⁡θ4\cdot \left( \dfrac{1-\sin \theta }{\cos \theta } \right)=\dfrac{1+\sin \theta }{\cos \theta }……………........( i )
On solving brackets,
4−4sin⁡θcos⁡θ=1+sin⁡θcos⁡θ\dfrac{4-4\sin \theta }{\cos \theta }=\dfrac{1+\sin \theta }{\cos \theta }
On simplifying equation, we get
4−4sin⁡θ=1+sin⁡θ4-4\sin \theta =1+\sin \theta
Taking, 4sin⁡θ4\sin \theta from left hand side to right hand side and 1 from right hand side to left hand side,
4−1=sin⁡θ+4sin⁡θ4-1=\sin \theta +4\sin \theta
On solving, we get
3=5sin⁡θ3=5\sin \theta
Taking, 4 from the numerator of right hand side to denominator of left hand side, we get
sin⁡θ=35\sin \theta =\dfrac{3}{5}
Hence, if sec⁡θ−tan⁡θsec⁡θ+tan⁡θ=14\dfrac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }=\dfrac{1}{4} , then the value of sin⁡θ\sin \theta is equals to 35\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.