Question & Answer
QUESTION

Solve the following equation to find the value of x: $\sin x\tan x-1=\tan x-\sin x$

ANSWER Verified Verified
Hint: First make the following substitution: $\tan x=\dfrac{\sin x}{\cos x}$. Then simplify LHS and RHS and cancel out the cos x in the denominator. Then rearrange and take the terms common to get $\left[ \sin x-1 \right]\left[ \sin x+\cos x \right]=0$. Solve for both the expressions: $\sin x-1=0$ and $\sin x+\cos x=0$ to find the final answer.

Complete step-by-step answer:

In this question, we need to solve the following equation to find the value of x: $\sin x\tan x-1=\tan x-\sin x$

We already know that tangent of an angle is the result of the division of sine of the angle and the cosine of that angle. i.e. $\tan x=\dfrac{\sin x}{\cos x}$

Using this property in the given equation, we will get the following:

$\sin x\tan x-1=\tan x-\sin x$

$\sin x\dfrac{\sin x}{\cos x}-1=\dfrac{\sin x}{\cos x}-\sin x$

$\dfrac{{{\sin }^{2}}x-\cos x}{\cos x}=\dfrac{\sin x\left[ 1-\cos x \right]}{\cos x}$

Now, we will take cos x common from the denominators of both the terms and then cancel it out. Doing this, we will get the following:

${{\sin }^{2}}x-\cos x=\sin x\left[ 1-\cos x \right]$

${{\sin }^{2}}x-\cos x=\sin x-\sin x\cos x$

Now, we will bring all the terms from the left hand side to the right hand side. Doing this, we will get the following:

${{\sin }^{2}}x-\sin x-\cos x+\sin x\cos x=0$

Now, we will take sin x common from the first two terms and we will take cos x common from the last two terms of the left hand side. Doing this, we will get the following:

$\sin x\left[ \sin x-1 \right]+\cos x\left[ \sin x-1 \right]=0$

Now, we have only two big terms left. So, we will take [sin x – 1] common from these two terms. Doing this, we will get the following:

$\left[ \sin x-1 \right]\left[ \sin x+\cos x \right]=0$

$\sin x-1=0$ or $\sin x+\cos x=0$

We know that $\sin x-1=0$ satisfies for $x=\left( 2n+1 \right)\dfrac{\pi }{2}$

We also know that $\sin x+\cos x=0$ satisfies for $x=\left( 4n-1 \right)\dfrac{\pi }{4}$

So, the solution of $\sin x\tan x-1=\tan x-\sin x$ is $x=\left( 2n+1 \right)\dfrac{\pi }{2},\left(

4n-1 \right)\dfrac{\pi }{4}$, where n is any whole number.

This is our final answer.

Note: In this question, it is very important to make the following transformation : $\tan x=\dfrac{\sin x}{\cos x}$ otherwise the simplification will get complicated. This way the expression is only in terms of sine and cosine which is easier to solve. Also, in the last few steps, remember to solve both the expressions: $\sin x-1=0$ and $\sin x+\cos x=0$. Many students solve just one of these and move on which is a partial answer.