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# Prove that trigonometric equation $\dfrac{\tan \theta }{\sec \theta +1}=\dfrac{\sec \theta -1}{\tan \theta }$.  Verified
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Hint: The given question is related to trigonometric identities. Try to recall the formulae related to the relationship between sine, cosine, tangent, and secant of an angle.

Before proceeding with the problem, first, let’s see the formulae used to solve the given problem.
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
$\sec \theta =\dfrac{1}{\cos \theta }$
$1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta$
We need to prove that $\dfrac{\tan \theta }{\sec \theta +1}=\dfrac{\sec \theta -1}{\tan \theta }$.
First, we will consider the left-hand side of the equation. The left-hand side of the equation is given as $\dfrac{\tan \theta }{\sec \theta +1}$ . We know $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$ . So, the left-hand side of the equation becomes $\dfrac{\dfrac{\sin \theta }{\cos \theta }}{\dfrac{1}{\cos \theta }+1}$ .
$\Rightarrow LHS=\dfrac{\dfrac{\sin \theta }{\cos \theta }}{\dfrac{1+\cos \theta }{\cos \theta }}$
$\Rightarrow LHS=\dfrac{\sin \theta }{1+\cos \theta }$
Now, we know that the value of a fraction does not change on multiplying and dividing the fraction by the same number, except $0$ .
So, $LHS=\dfrac{\sin \theta }{1+\cos \theta }\times \dfrac{1-\cos \theta }{1-\cos \theta }$ .
$\Rightarrow LHS=\dfrac{\sin \theta \left( 1-\cos \theta \right)}{1-{{\cos }^{2}}\theta }$
Now, we know that the value of $1-{{\cos }^{2}}\theta$ is equal to ${{\sin }^{2}}\theta$. So, the value of the left-hand side of the equation becomes $\dfrac{\sin \theta \left( 1-\cos \theta \right)}{{{\sin }^{2}}\theta }$ .
$\Rightarrow LHS=\dfrac{1-\cos \theta }{\sin \theta }$
So, the value of the left-hand side of the equation is equal to $\dfrac{1-\cos \theta }{\sin \theta }$ .
Now, we will consider the right-hand side of the equation. The right-hand side of the equation is given as $\dfrac{\sec \theta -1}{\tan \theta }$ . We know $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$ . So, the right-hand side of the equation becomes $\dfrac{\dfrac{1}{\cos \theta }-1}{\dfrac{\sin \theta }{\cos \theta }}$ .
$\Rightarrow RHS=\dfrac{\dfrac{1-\cos \theta }{\cos \theta }}{\dfrac{\sin \theta }{\cos \theta }}$
$\Rightarrow RHS=\dfrac{1-\cos \theta }{\sin \theta }$
So, the value of the right-hand side of the equation is equal to $\dfrac{1-\cos \theta }{\sin \theta }$ . Clearly, the values of the left-hand side of the equation and the right-hand side of the equation are the same, i.e. LHS=RHS. Hence, proved.

Note: Students generally get confused and write $1+{{\cos }^{2}}\theta ={{\sin }^{2}}\theta$ instead of $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta$ , which is wrong. These formulae should be properly remembered without any mistake as confusion in the formula can result in getting a wrong answer.