# Prove that trigonometric equation $\dfrac{\tan \theta }{\sec \theta +1}=\dfrac{\sec \theta -1}{\tan \theta }$.

Last updated date: 20th Mar 2023

•

Total views: 304.5k

•

Views today: 3.87k

Answer

Verified

304.5k+ views

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE