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Chapter: Trigonometry· Mathematics Part 2 (Geometry)

Asked in: July 2025 board paperIf \(\tan\theta = 1\), then find the value of \(\dfrac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta}\) by completing the following activity.Activity: \(\tan\theta = 1\) (given), but \(\tan\text{____}…

Asked in: July 2025 board paper

If \(\tan\theta = 1\), then find the value of \(\dfrac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta}\) by completing the following activity.

Activity: \(\tan\theta = 1\) (given), but \(\tan\text{____} = 1\) \(\therefore \theta = \text{____}\)

\(\therefore \dfrac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta} = \dfrac{\sin 45^\circ + \cos 45^\circ}{\sec 45^\circ + \operatorname{cosec} 45^\circ} = \dfrac{\dfrac{1}{\text{____}} + \dfrac{1}{\sqrt2}}{\sqrt2 + \text{____}} = \dfrac{\dfrac{2}{\sqrt2}}{\text{____}} = \dfrac{1}{\text{____}}\)

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Answer: \(\dfrac{1}{2}\)

\(\tan 45^\circ = 1\) \(\therefore \theta = 45^\circ\).

\(\sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt2}\) and \(\sec 45^\circ = \operatorname{cosec} 45^\circ = \sqrt2\).

\(\dfrac{\dfrac{1}{\sqrt2} + \dfrac{1}{\sqrt2}}{\sqrt2 + \sqrt2} = \dfrac{\dfrac{2}{\sqrt2}}{2\sqrt2} = \dfrac{2}{\sqrt2 \times 2\sqrt2} = \dfrac{2}{4} = \dfrac{1}{2}\)

Blanks: 45°; 45°; \(\sqrt2\); \(\sqrt2\); \(2\sqrt2\); 2.

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