Question

Prove that \[\dfrac{{\sec x + 1 - \tan x}}{{\tan x + \sec x + 1}} = \dfrac{{1 - \sin x}}{{\cos x}}\]

Answer 1

Hint:
Here, we are required to prove that the given LHS is equal to the given RHS. We will use the trigonometric identity of the sum of squares of tangent and secant and substitute it in the place of 1 in the numerator. Hence, after taking common, canceling out the same terms, and using the appropriate identities we will be able to prove that the given two sides are equal.

Formula Used:
We will use the following formulas:
1) \[{\sec ^2}x - {\tan ^2}x = 1\]
2) \[\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)\]

Complete Step by step Solution:
We have to \[\dfrac{{\sec x + 1 - \tan x}}{{\tan x + \sec x + 1}} = \dfrac{{1 - \sin x}}{{\cos x}}\].
We will first take into consideration the left hand side of the equation and solve it further.
LHS \[ = \dfrac{{\sec x + 1 - \tan x}}{{\tan x + \sec x + 1}}\]
Now, we know the trigonometric formula: \[{\sec ^2}x - {\tan ^2}x = 1\]
Hence, in the numerator of the LHS, we will substitute the value of 1 using the above trigonometric formula.
\[ \Rightarrow \] LHS \[ = \dfrac{{\sec x - \tan x + \left( {{{\sec }^2}x - {{\tan }^2}x} \right)}}{{\tan x + \sec x + 1}}\]
Now, using the identity \[\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)\], we get
\[ \Rightarrow \] LHS \[ = \dfrac{{\left( {\sec x - \tan x} \right) + \left( {\sec x - \tan x} \right)\left( {\sec x + \tan x} \right)}}{{\tan x + \sec x + 1}}\]
Now, we will take \[\left( {\sec x - \tan x} \right)\] common from the numerator. Therefore, we get
\[ \Rightarrow \] LHS \[ = \dfrac{{\left( {\sec x - \tan x} \right)\left[ {1 + \sec x + \tan x} \right]}}{{\tan x + \sec x + 1}}\]
As we can see, one of the brackets in the numerator is the same as the denominator.
Cancelling out the common terms, we get
\[ \Rightarrow \] LHS \[ = \left( {\sec x - \tan x} \right)\]
Now, we know that, \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] and \[\sec x = \dfrac{1}{{\cos x}}\]
Therefore, substituting these values in the above equation, we get
\[ \Rightarrow \] LHS \[ = \dfrac{1}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}}\]
\[ \Rightarrow \] LHS \[ = \dfrac{{1 - \sin x}}{{\cos x}}\]
\[ \Rightarrow \] LHS \[ = \] RHS
Hence, proved

Therefore, we have proved that \[\dfrac{{\sec x + 1 - \tan x}}{{\tan x + \sec x + 1}} = \dfrac{{1 - \sin x}}{{\cos x}}\]

Note:
This question involved Trigonometry which is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, cartographers (to make maps) use trigonometry. It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine and the cosine function. In the simple terms they are written as ‘sin’, ‘cos’ and ‘tan’.