If cot θ = \(\dfrac{7}{8}\), evaluate:
(i) \(\dfrac{(1 + \sin θ)(1 - \sin θ)}{(1 + \cos θ)(1 - \cos θ)}\)
(ii) cot² θ
(i) \(\dfrac{49}{64}\)
(ii) \(\dfrac{49}{64}\)
Given: \(\cot \theta = \dfrac{7}{8}\)
To find:
(i) \(\dfrac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}\)
(ii) \(\cot^2 \theta\)
Step 1: Form a right triangle using \(\cot\theta\)
In a right-angled triangle (with angle \(\theta\)),
\(\cot\theta = \dfrac{\text{adjacent}}{\text{opposite}}\).
So take adjacent side = 7 units and opposite side = 8 units.
Step 2: Find the hypotenuse using Pythagoras theorem
Let hypotenuse be \(h\). Then
\(h^2 = 7^2 + 8^2 = 49 + 64 = 113\)
\(\Rightarrow h = \sqrt{113}\).
Step 3: Write \(\sin\theta\) and \(\cos\theta\) using sides
\(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{8}{\sqrt{113}}\)
\(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{7}{\sqrt{113}}\)
Step 4: Evaluate part (i)
First expand each product:
\((1+\sin\theta)(1-\sin\theta) = 1 - (\sin\theta)^2\)
\((1+\cos\theta)(1-\cos\theta) = 1 - (\cos\theta)^2\)
Now substitute \(\sin\theta = \dfrac{8}{\sqrt{113}}\) and \(\cos\theta = \dfrac{7}{\sqrt{113}}\):
Numerator:
\(1 - (\sin\theta)^2 = 1 - \left(\dfrac{8}{\sqrt{113}}\right)^2 = 1 - \dfrac{64}{113} = \dfrac{49}{113}\)
Denominator:
\(1 - (\cos\theta)^2 = 1 - \left(\dfrac{7}{\sqrt{113}}\right)^2 = 1 - \dfrac{49}{113} = \dfrac{64}{113}\)
Therefore,
\(\dfrac{(1 + \sin\theta)(1 - \sin\theta)}{(1 + \cos\theta)(1 - \cos\theta)} = \dfrac{\frac{49}{113}}{\frac{64}{113}} = \dfrac{49}{64}\)
So, (i) = \(\dfrac{49}{64}\).
Step 5: Evaluate part (ii)
\(\cot^2\theta = \left(\dfrac{7}{8}\right)^2 = \dfrac{49}{64}\).
Final Answers:
(i) \(\dfrac{49}{64}\)
(ii) \(\dfrac{49}{64}\)