मराठी

If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = ______.

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प्रश्न

If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = ______.

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उत्तर

If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = `pi/4`.

Explanation:

Given that 3 tan (θ – 15°) = tan (θ + 15°)

Which can be rewritten as `(tan(theta + 15^circ))/(tan(theta - 15^circ)) = 3/1`

Applying componendo and Dividendo

We get `(tan(theta + 15^circ) + tan(theta - 15^circ))/(tan(theta + 15^circ) - tan(theta - 15^circ))` = 2

⇒ `(sin(theta + 15^circ) cos(theta - 15^circ) + sin(theta - 15^circ) cos(theta + 15^circ))/(sin(theta + 15^circ) cos(theta - 15^circ) - sin(theta - 15^circ) cos(theta + 15^circ))` = 2

⇒ `(sin 2theta)/(sin30^circ)` = 2

i.e., sin 2θ = 1

Giving θ = `pi/4`

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पाठ 3: Trigonometric Functions - Solved Examples [पृष्ठ ५०]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 3 Trigonometric Functions
Solved Examples | Q 20 | पृष्ठ ५०

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