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Question: If \(3\cos \theta -4\sin \theta =2\cos \theta +\sin \theta \), is given, then find the value of \(\t...

If 3cos⁡θ−4sin⁡θ=2cos⁡θ+sin⁡θ3\cos \theta -4\sin \theta =2\cos \theta +\sin \theta , is given, then find the value of tan⁡θ\tan \theta .

Explanation

Solution

Hint:In order to find the solution of this question, we will start from the given equation and then we will try to form tan⁡θ\tan \theta , by using the formula, tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and then we will calculate the value of tan⁡θ\tan \theta .

Complete step-by-step answer:
In this question, we have been asked to find the value of tan⁡θ\tan \theta , when it is given that 3cos⁡θ−4sin⁡θ=2cos⁡θ+sin⁡θ3\cos \theta -4\sin \theta =2\cos \theta +\sin \theta . To solve this, we will first consider the given equality, that is, 3cos⁡θ−4sin⁡θ=2cos⁡θ+sin⁡θ3\cos \theta -4\sin \theta =2\cos \theta +\sin \theta . We will try to form sin⁡θcos⁡θ\dfrac{\sin \theta }{\cos \theta } here. For that, we will write the terms of cos⁡θ\cos \theta on the left hand side and the terms of sin⁡θ\sin \theta on the right hand side. So, we can write the given equation 3cos⁡θ−4sin⁡θ=2cos⁡θ+sin⁡θ3\cos \theta -4\sin \theta =2\cos \theta +\sin \theta as,
3cos⁡θ−2cos⁡θ=sin⁡θ+4sin⁡θ3\cos \theta -2\cos \theta =\sin \theta +4\sin \theta
We know that arithmetic operations are applied to like terms. So, we get the above equation as,
cos⁡θ=5sin⁡θ\cos \theta =5\sin \theta
Now, we will divide the whole equation by cos⁡θ\cos \theta . On doing so, we get,
cos⁡θcos⁡θ=5sin⁡θcos⁡θ\dfrac{\cos \theta }{\cos \theta }=\dfrac{5\sin \theta }{\cos \theta }
We know that the common terms of the numerator and the denominator gets cancelled, so we get,
1=5sin⁡θcos⁡θ1=\dfrac{5\sin \theta }{\cos \theta }
We will now divide the equation by 5. So, we get,
15=5sin⁡θ5cos⁡θ ⇒sin⁡θcos⁡θ=15 \begin{aligned} & \dfrac{1}{5}=\dfrac{5\sin \theta }{5\cos \theta } \\\ & \Rightarrow \dfrac{\sin \theta }{\cos \theta }=\dfrac{1}{5} \\\ \end{aligned}
We also know that sin⁡θcos⁡θ\dfrac{\sin \theta }{\cos \theta } can be expressed as tan⁡θ\tan \theta . So, applying that, we get,
tan⁡θ=15\tan \theta =\dfrac{1}{5}
Hence, we can say that tan⁡θ=15\tan \theta =\dfrac{1}{5}, if 3cos⁡θ−4sin⁡θ=2cos⁡θ+sin⁡θ3\cos \theta -4\sin \theta =2\cos \theta +\sin \theta .

Note: While solving the question, the possible mistakes that can be made are in the calculations. Also, we have to remember that sin⁡θcos⁡θ\dfrac{\sin \theta }{\cos \theta } can be expressed as tan⁡θ\tan \theta . So, we should try to form tan⁡θ\tan \theta , either by taking the terms to one side and then dividing them by cos⁡θ\cos \theta .