Question
Prove the following trigonometric identities.
$\frac{1+\cos\text{A}}{\sin\text{A}}=\frac{\sin\text{A}}{1-\cos\text{A}}$
$\frac{1+\cos\text{A}}{\sin\text{A}}=\frac{\sin\text{A}}{1-\cos\text{A}}$
✓
Answer
We need to prove $\frac{1+\cos\text{A}}{\sin\text{A}}=\frac{\sin\text{A}}{1-\cos\text{A}}$
Now, multiplying the numerator and denominator of L.H.S by $1-\cos\text{A}$, we get
$\frac{1+\cos\text{A}}{\sin\text{A}}=\frac{1+\cos\text{A}}{\sin\text{A}}\times\frac{1-\cos\text{A}}{1-\cos\text{A}}$
Further using the identity $a^2 - b^2 = (a + b)(a - b)$, we get
$\text{L.H.S}=\frac{1+\cos\text{A}}{\sin\text{A}}\times\frac{1-\cos\text{A}}{1-\cos\text{A}}=\frac{1-\cos^2\text{A}}{\sin\text{A}(1-\cos\text{A})}$
$=\frac{\sin^2\text{A}}{\sin\text{A}(1-\cos\text{A})} \ (\text{using} \sin^2\theta+\cos^2\theta=1)$
$=\frac{\sin\text{A}}{1-\cos\text{A}}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Now, multiplying the numerator and denominator of L.H.S by $1-\cos\text{A}$, we get
$\frac{1+\cos\text{A}}{\sin\text{A}}=\frac{1+\cos\text{A}}{\sin\text{A}}\times\frac{1-\cos\text{A}}{1-\cos\text{A}}$
Further using the identity $a^2 - b^2 = (a + b)(a - b)$, we get
$\text{L.H.S}=\frac{1+\cos\text{A}}{\sin\text{A}}\times\frac{1-\cos\text{A}}{1-\cos\text{A}}=\frac{1-\cos^2\text{A}}{\sin\text{A}(1-\cos\text{A})}$
$=\frac{\sin^2\text{A}}{\sin\text{A}(1-\cos\text{A})} \ (\text{using} \sin^2\theta+\cos^2\theta=1)$
$=\frac{\sin\text{A}}{1-\cos\text{A}}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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