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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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.

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