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14. Prove that:
[tex]\[ \frac{1-\sin \theta}{\cos \theta} = \frac{\cos \theta}{1+\sin \theta} \][/tex]

Sagot :

GinnyAnsweridn
Sure, let's prove the given trigonometric identity:
[tex]\[ \frac{1 - \sin \theta}{\cos \theta} = \frac{\cos \theta}{1 + \sin \theta} \][/tex]

### Step-by-Step Solution:

#### Step 1: Analyze the Left-Hand Side (LHS)
The left-hand side (LHS) of the equation is:
[tex]\[ \frac{1 - \sin \theta}{\cos \theta} \][/tex]

#### Step 2: Simplify the LHS
Let's simplify the LHS using trigonometric identities:

One useful strategy is to express the terms in a different form:
[tex]\[ \frac{1 - \sin \theta}{\cos \theta} = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \sec \theta - \tan \theta \][/tex]

However, let's instead further simplify and check:

#### Step 3: Analyze the Right-Hand Side (RHS)
The right-hand side (RHS) of the equation is:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \][/tex]

#### Step 4: Simplify the RHS
Let's simplify the RHS using trigonometric identities. You can see it as it is already simplified:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \][/tex]

#### Step 5: Compare the Simplified Forms of LHS and RHS
Now, we see that we have:
[tex]\[ \sec \theta - \tan \theta \][/tex]
and
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \][/tex]

To prove these two simplified forms are equal, let's see if the left-hand side matches the right-hand side upon further simplification.

#### Step 6: Attempt to Show Equivalence
We assume our identity:
[tex]\[ \frac{1 - \sin \theta}{\cos \theta} = \frac{\cos \theta}{1 + \sin \theta} \][/tex]

To show these are equal, we cross-multiply to get a common expression to compare both sides:
[tex]\[ (1 - \sin \theta)(1 + \sin \theta) = \cos^2 \theta \][/tex]

The left side simplifies using difference of squares:
[tex]\[ 1 - \sin^2 \theta \][/tex]

Using the Pythagorean identity:
[tex]\[ \cos^2 \theta \][/tex]

So, both sides leave us with:
[tex]\[ \cos^2 \theta = \cos^2 \theta \][/tex]

This confirms that:
[tex]\[ \frac{1 - \sin \theta}{\cos \theta} = \frac{\cos \theta}{1 + \sin \theta} \][/tex]

#### Conclusion
We have shown the initial expression holds true. Thus, the given trigonometric identity:
[tex]\[ \frac{1 - \sin \theta}{\cos \theta} = \frac{\cos \theta}{1 + \sin \theta} \][/tex]
is indeed valid and proved.
Dashataranidn

Solution:

[tex]\text{L.H.S. $=\dfrac{1-\sin\theta}{\cos\theta}=\dfrac{1}{\cos\theta}-\dfrac{\sin\theta}{\cos\theta}=\sec\theta-\tan\theta$}[/tex]

[tex]\text{R.H.S. $=\dfrac{\cos\theta}{1+\sin\theta}$}\\\\[/tex]

           [tex]=\dfrac{\cos\theta}{1+\sin\theta}\times\dfrac{1-\sin\theta}{1-\sin\theta}[/tex]

           [tex]=\dfrac{\cos\theta-\cos\theta\sin\theta}{1-\sin^2\theta}[/tex]

           [tex]=\dfrac{\cos\theta-\cos\theta\sin\theta}{\cos^2\theta}[/tex]

           [tex]=\dfrac{\cos\theta}{\cos^2\theta}-\dfrac{\cos\theta\sin\theta}{\cos^2\theta}[/tex]

           [tex]=\dfrac{1}{\cos\theta}-\dfrac{\sin\theta}{\cos\theta}[/tex]

           [tex]=\sec\theta-\tan\theta[/tex]

[tex]\therefore\ \text{L.H.S. = R.H.S., proved$}[/tex]

         

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