Mathematics

Mathematics

Problem

7. If cot⁡θ=78\cot \theta=\frac{7}{8}, evaluate : (i) (1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}, (ii) cot⁡2θ\cot ^{2} \theta (i) (1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} (ii) cot⁡2θ\cot ^{2} \theta
Intuitive explanation

You're working with a cotangent ratio and need to extract information about the trig functions themselves. The key insight is that cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta} connects the two functions, and the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 lets you find their individual values. Once you know sin⁡θ\sin \theta and cos⁡θ\cos \theta, both expressions become straightforward algebra—watch how the numerators and denominators simplify using difference-of-squares patterns.

Intuitive explanation

You're working with a cotangent ratio and need to extract information about the trig functions themselves. The key insight is that cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta} connects the two functions, and the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 lets you find their individual values. Once you know sin⁡θ\sin \theta and cos⁡θ\cos \theta, both expressions become straightforward algebra—watch how the numerators and denominators simplify using difference-of-squares patterns.

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