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Question: How do you prove \(\dfrac{1}{{1 + \sin \left( \theta \right)}} + \dfrac{1}{{1 - \sin \left( \theta \...

How do you prove 11+sin⁡(θ)+11−sin⁡(θ)=2sec⁡2(θ)\dfrac{1}{{1 + \sin \left( \theta \right)}} + \dfrac{1}{{1 - \sin \left( \theta \right)}} = 2{\sec ^2}\left( \theta \right)?

Explanation

Solution

In order to prove the given expression, we will first consider the complex side of the equation, rewrite it as a single term, and then, simplify it using the different identities and formulae. Through successful application of various identities and formulae, we simplify the expression until we get the same expression as on the other side of the equation.

Complete step-by-step solution:
We need to prove: 11+sin⁡(θ)+11−sin⁡(θ)=2sec⁡2(θ)\dfrac{1}{{1 + \sin \left( \theta \right)}} + \dfrac{1}{{1 - \sin \left( \theta \right)}} = 2{\sec ^2}\left( \theta \right)
Proof:
First, we consider the complex side of the equation, i.e., the LHS:
11+sin⁡(θ)+11−sin⁡(θ)\dfrac{1}{{1 + \sin \left( \theta \right)}} + \dfrac{1}{{1 - \sin \left( \theta \right)}}
Rewriting the above expression as a single term, we have:
=1−sin⁡(θ)+[1+sin⁡(θ)][1+sin⁡(θ)][1−sin⁡(θ)]= \dfrac{{1 - \sin \left( \theta \right) + \left[ {1 + \sin \left( \theta \right)} \right]}}{{\left[ {1 + \sin \left( \theta \right)} \right]\left[ {1 - \sin \left( \theta \right)} \right]}} ………………………….(1)
We know that: (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
Thus, the denominator of the expression (1) can be rewritten as:
[1+sin⁡(θ)][1−sin⁡(θ)]=1−sin⁡2(θ)\left[ {1 + \sin \left( \theta \right)} \right]\left[ {1 - \sin \left( \theta \right)} \right] = 1 - {\sin ^2}\left( \theta \right) ……………………….(2)
Therefore, after substituting the value of [1+sin⁡(θ)][1−sin⁡(θ)]\left[ {1 + \sin \left( \theta \right)} \right]\left[ {1 - \sin \left( \theta \right)} \right] from equation (2) in expression (1), and also simplifying the numerator, we get:
=1−sin⁡(θ)+1+sin⁡(θ)1−sin⁡2(θ)= \dfrac{{1 - \sin \left( \theta \right) + 1 + \sin \left( \theta \right)}}{{1 - {{\sin }^2}\left( \theta \right)}}
On simplifying further, we get:
=21−sin⁡2(θ)= \dfrac{2}{{1 - {{\sin }^2}\left( \theta \right)}} ……………………….(3)
We know the trigonometric identity: sin⁡2(θ)+cos⁡2(θ)=1{\sin ^2}\left( \theta \right) + {\cos ^2}\left( \theta \right) = 1 ……………………..(4)
On rearranging equation (4), we get:
1−sin⁡2(θ)=cos⁡2(θ)1 - {\sin ^2}\left( \theta \right) = {\cos ^2}\left( \theta \right) ……………………..(5)
Now, substituting the value of [1−sin⁡2(θ)]\left[ {1 - {{\sin }^2}\left( \theta \right)} \right] from equation (5) in the expression (3), we get:
=2cos⁡2(θ)= \dfrac{2}{{{{\cos }^2}\left( \theta \right)}} …………………………(6)
Again, from trigonometric identities:
1cos⁡θ=sec⁡θ⇒1cos⁡2θ=sec⁡2θ\dfrac{1}{{\cos \theta }} = \sec \theta \Rightarrow \dfrac{1}{{{{\cos }^2}\theta }} = {\sec ^2}\theta ………………………….(7)
Thus, substituting the respective value from equation (7) in the expression (6), we finally have our expression as:
=2sec⁡2(θ)= 2{\sec ^2}\left( \theta \right), which is nothing but the expression on the other side of the expression, that is, on the right side of the expression.
Hence it is proved

Note: To prove a trigonometric equation, we should tend to start with the more complicated side of the equation, and keep simplifying it until it is transformed into the same expression as on the other side of the given equation. In some cases, we can also try to simplify both sides of the equation and arrive at a common expression to prove their equality. The various procedures of solving a trigonometric equation are: expanding the expressions, making use of the identities, factoring the expressions or simply using basic algebraic strategies to obtain the desired results.