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If tan A = 1 - cos B , then tan 2A is equal to sin B
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- cot B
- tan B
- cos B
- cosec B
- cot B
Correct Option: B
| Here, tanA = | ||
| sinB |
We know that,
| tan2A = | ||
| 1 - tan²A |
| tan2A = | 2 |  |  | |
| sinB | ||||
| 1 - | | | ² | |
| sinB | ||||
| tan2A = | |||
| sinB | |||
| sin²B | |||
| = | [∵ sin²θ = 1 - cos²θ] | |
| 1 - cos²B - (1 - cosB)² |
| = | ||
| (1 - cosB)[1 + cosB - 1 + cosB] |
| = | ||
| 2cosB |
= tan B.
