If \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\] , then find the value of tan2A.

Given: \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\] \[ \Rightarrow \text{ tan } A = \frac{2 \sin^2 \frac{B}{2}}{2\sin\frac{B}{2}\cos\frac{B}{2}} \left( 1 - \cos2\theta = 2 \sin^2 \theta \text{ and } \sin2\theta = 2\sin\theta\cos\theta \right)\] \[ \Rightarrow \text{ tan } A = \frac{\sin\frac{B}{2}}{\cos\frac{B}{2}} = \tan\frac{B}{2}\] \[ \Rightarrow A = \frac{B}{2}\] \[\Rightarrow 2A = B\]\[ \therefore \tan2A = \text{ tan } B\] Hence, the value of tan2A is tanB.

If Tan a = 1 − Cos B Sin B , Then Find the Value of Tan2a. - Mathematics

प्रश्न

If $tan A = \frac{1 - cos B}{sin B}$

, then find the value of tan2A.

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उत्तर

Given:

$tan A = \frac{1 - cos B}{sin B}$

$\Rightarrow tan A = \frac{2 \sin^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}} (1 - \cos 2 θ = 2 \sin^{2} θ and \sin 2 θ = 2 \sin θ \cos θ)$

$\Rightarrow tan A = \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}} = \tan \frac{B}{2}$

$\Rightarrow A = \frac{B}{2}$

$\Rightarrow 2 A = B$
$∴ \tan 2 A = tan B$

Hence, the value of tan2A is tanB.

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 11 Q 10 Q 12

पाठ 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.4 [पृष्ठ ४२]

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आरडी शर्मा Mathematics [English] Class 11

पाठ 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.4 | Q 11 | पृष्ठ ४२

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If Tan a = 1 − Cos B Sin B , Then Find the Value of Tan2a. - Mathematics

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If Tan a = 1 − Cos B Sin B , Then Find the Value of Tan2a. - Mathematics

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