Question

Prove that sec(3π/2).sec(θ - 5π/2) + tan(5π/2 + θ ).tan(θ - 3π/2) = (-1)

Answer 1

Answer on Question #79261 – Math – Trigonometry

Question

Prove that $\sec \left(\frac{3\pi}{2} - \theta\right) \cdot \sec \left(\theta - \frac{5\pi}{2}\right) + \tan \left(\frac{5\pi}{2} + \theta\right) \cdot \tan \left(\theta - \frac{3\pi}{2}\right) = (-1)$

Solution

$$\begin{aligned} \text{L.H.S} &= \sec \left(\frac{3\pi}{2} - \theta\right) \cdot \sec \left(\theta - \frac{5\pi}{2}\right) + \tan \left(\frac{5\pi}{2} + \theta\right) \cdot \tan \left(\theta - \frac{3\pi}{2}\right) \\ &= \sec \left(\frac{3\pi}{2} - \theta\right) \cdot \sec \left[ -\left(\frac{5\pi}{2} - \theta\right) \right] + \tan \left(\frac{5\pi}{2} + \theta\right) \cdot \tan \left[ -\left(\frac{3\pi}{2} - \theta\right) \right] \\ &= \sec \left(\frac{3\pi}{2} - \theta\right) \cdot \sec \left(\frac{5\pi}{2} - \theta\right) + \tan \left(\frac{5\pi}{2} + \theta\right) \cdot \left[ -\tan \left(\frac{3\pi}{2} - \theta\right) \right] \\ &= \left[ -\text{cosec } \theta \right] \cdot \left[ \text{cosec } \theta \right] + \left[ -\text{cot } \theta \right] \cdot \left[ -\text{cot } \theta \right] \\ &= -\text{cosec}^2 \theta + \text{cot}^2 \theta \\ &= -1 \quad \left[ \text{As, cosec}^2 \theta - \text{cot}^2 \theta = \frac{1}{\sin^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1 - \cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta} = 1 \right] \\ &= \text{R.H.S} \end{aligned}$$

Answer: $\sec \left(\frac{3\pi}{2} - \theta\right) \cdot \sec \left(\theta - \frac{5\pi}{2}\right) + \tan \left(\frac{5\pi}{2} + \theta\right) \cdot \tan \left(\theta - \frac{3\pi}{2}\right) = -1$ has been proved.

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