Mathematics — Standard (041) — CBSE Class 10 board question

Given: $\sin\theta + \cos\theta = \sqrt{3}$

To prove: $\tan\theta + \cot\theta = 1$

Squaring both sides of the given equation:

$$(\sin\theta + \cos\theta)^2 = (\sqrt{3})^2$$

$$\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 3$$

Since $\sin^2\theta + \cos^2\theta = 1$:

$$1 + 2\sin\theta\cos\theta = 3$$

$$2\sin\theta\cos\theta = 2$$

$$\sin\theta\cos\theta = 1$$

Now, LHS $= \tan\theta + \cot\theta = \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}$

$$= \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{1} = 1 = \text{RHS}$$

Hence proved.

Source: Chapter 8, Trigonometric Identities

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• The key move is squaring the given condition to extract $\sin\theta\cos\theta$.
• Use the identity $\sin^2\theta + \cos^2\theta = 1$ twice — once after squaring, and once in the numerator of $\tan\theta + \cot\theta$.
• Examiners award 1 mark for squaring and simplifying to get $\sin\theta\cos\theta = 1$, 1 mark for correct expansion of $\tan\theta + \cot\theta$, and 1 mark for the final step leading to the result.