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Question: How do you find the identity \(\dfrac{{\cos t}}{{1 - \sin t}} = \sec t + \tan t\)?...

How do you find the identity cos⁡t1−sin⁡t=sec⁡t+tan⁡t\dfrac{{\cos t}}{{1 - \sin t}} = \sec t + \tan t?

Explanation

Solution

To prove this trigonometric identity we will use the concept of solving the left hand side and right hand side separately so to get the same value in both the sides. So, for the left hand side, we will first rationalise the function with the appropriate function in order to simplify the function so that we can eventually get the form of the right hand side. Further we will use the required operations and use required trigonometric identities.Then, we will equate the right hand side in a similar manner if required. So, let us see how to solve this problem.

Complete step by step answer:
To prove, the given identity is cos⁡t1−sin⁡t=sec⁡t+tan⁡t\dfrac{{\cos t}}{{1 - \sin t}} = \sec t + \tan t. First operating the left hand side,
⇒cos⁡t1−sin⁡t\Rightarrow \dfrac{{\cos t}}{{1 - \sin t}}
Rationalising the function by multiplying both numerator and denominator by (1+sin⁡t)\left( {1 + \sin t} \right), we get,
⇒cos⁡t(1+sin⁡t)(1−sin⁡t)(1+sin⁡t)\Rightarrow \dfrac{{\cos t\left( {1 + \sin t} \right)}}{{(1 - \sin t)\left( {1 + \sin t} \right)}}

We know, (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}.
So, using this identity in the above function, we get,
⇒cos⁡t(1+sin⁡t)(1−sin⁡2t)\Rightarrow \dfrac{{\cos t\left( {1 + \sin t} \right)}}{{(1 - {{\sin }^2}t)}}
Now, we know the trigonometric identity, cos⁡2x+sin⁡2x=1{\cos ^2}x + {\sin ^2}x = 1.
Subtracting sin⁡2x{\sin ^2}x from both the sides of the identity, we get,
⇒cos⁡2x=1−sin⁡2x\Rightarrow {\cos ^2}x = 1 - {\sin ^2}x
Now, using this identity in the given function, we get,
⇒cos⁡t(1+sin⁡t)cos⁡2t\Rightarrow \dfrac{{\cos t\left( {1 + \sin t} \right)}}{{{{\cos }^2}t}}

Now, simplifying the function, we get,
⇒(1+sin⁡t)cos⁡t\Rightarrow \dfrac{{\left( {1 + \sin t} \right)}}{{\cos t}}
Now, we can write it as,
⇒1cos⁡t+sin⁡tcos⁡t\Rightarrow \dfrac{1}{{\cos t}} + \dfrac{{\sin t}}{{\cos t}}
Now, we know, 1cos⁡t=sec⁡t\dfrac{1}{{\cos t}} = \sec t and sin⁡tcos⁡t=tan⁡t\dfrac{{\sin t}}{{\cos t}} = \tan t.
Using these trigonometric identities in the given function, we get,
⇒sec⁡t+tan⁡t\Rightarrow \sec t + \tan t
Now, right hand side is,
⇒sec⁡t+tan⁡t\Rightarrow \sec t + \tan t
Therefore, Left Hand Side = Right Hand Side

Hence, cos⁡t1−sin⁡t=sec⁡t+tan⁡t\dfrac{{\cos t}}{{1 - \sin t}} = \sec t + \tan t.

Note: We can also solve the problem in the opposite direction also, that is, we can operate the right hand side first and then the left hand side. We could have written secant and tangent function in the right hand side in the form of cosine and sine function and then operate accordingly to equate it to the left hand side. It's always easier to operate with sine and cosine functions instead of other trigonometric functions.